Unit VI Molarity

Lesson Day Date Topic

1. Molarity 1

2. Molarity Lab Molarity 2

3. Stoichiometry

4. Titrations lab 1 Homework WS # 4

5. Titrations Lab 2 Homework WS # 5

6. Dilutions

7. Spectrophotometry Lab

8. Molarity and Dilutions

9. Ion Concentration

10. Molarity Unit Review # 1

11. Molarity Unit Review # 2

12. Chemistry 11 Calculations Practice Test # 1

13. Chemistry 11 Calculations Practice Test # 2

Molarity Worksheet # 1

1. 15.8 g of KCl is dissolved in 225 mL of water. Calculate the molarity.

15.8 g x 1 mole

Molarity = 74.6 g = 0.941 M

0.225 L

2. Calculate the mass of KCl required to prepare 250. mL of 0.250 M solution.

0.250 L x 0.250 moles x 74.6 g = 4.66 g

1 L 1 mole

3. Calculate the volume of 0.30 M KCl solution that contains 6.00 g of KCl.

6.00 g x 1 mole x 1 L = 0.27 L

74.6 g 0.30 mol

4. Calculate the volume of 0.250 M H₂SO₄ that contains 0.250 g H₂SO₄.

0.250 g H₂SO₄ x 1 mole x 1 L = 0.0102 L

98.12 g 0.250 mole

5. 1.50 g of NaCl is dissolved in 100.0 mL of water. Calculate the concentration.

1.50 g x 1 mole

Molarity = 58.5 g = 0.256 M

0.1000 L

6. How many moles of NaCl are in 250. mL of a 0.200 M solution?

0.250 L x 0.200 mole = 0.0500 moles

1 L

7. How many litres of a 0.200 M KCl solution contain 0.250 moles?

0.250 moles x 1 L = 1.25 L

0.200 moles

8. How many millilitres of 0.200 M H₂SO₄ are required to completely neutralize 250. mL of 0.250 M NaOH?

H₂SO₄ + 2NaOH → Na₂SO₄ + 2HOH

? mL 0.250 L

0.250 L NaOH x 0.250 mole x 1 mole H₂SO₄ x 1 L x 1000 mL = 156 mL

1 L 2 mole NaOH 0.200 mole 1 L

9. Calculate the mass of CuSO₄^.5H₂O required to prepare 100.0 mL of 0.100 M solution.

0.100 L x 0.100 mole x 249.7 g = 2.50 g

1 L 1 mole

10.     Calculate the mass of Cu(NO₃)₂^.6H₂O required to prepare 100.0 mL of 0.200 M solution.

5.91 g

11.     Calculate the mass of CoCl₃^.6H₂O required to prepare 500.0 mL of a 0.200 M       solution.

27.4 g

12.     50.0 g of NaCl is dissolved in 200.0 mL of water, calculate the molarity.

       4.27 M

13.     25.0 g of CuSO₄^.8H₂O is dissolved in 25.0 mL of water, calculate the molarity.

       3.29 M

14.     Calculate the mass of NaCl required to prepare 500.0 mL of a 0.500 M solution.

       14.6 g

15.     Calculate the volume of 0.500 M NaCl solution required to contain 0.0500 g of NaCl.

0.00171 L

16.     Calculate the volume of 0.200 M NaCl solution required to contain

          0.653 g of NaCl.

0.0558L

17.     Calculate the mass of NaCl required to prepare 256 mL of a 0.35 M solution.

5.2 g

18.     25.2 g of NaCl is dissolved in 365 mL of water, calculate the molarity.

1.18 M

19.     56.3 g of CuSO₄^.8H₂O is dissolved in 30. mL of water, calculate the molarity.

6.2 M

Worksheet # 2 Molarity

1. Calculate the mass of CuSO₄^.6H₂O required to prepare 200.0 mL of 0.300 M solution.

0.200 L x 0.300 moles x 267.72 g = 16.1 g

1 L 1 mole

2. Calculate the mass of CoCl₃^.8H₂O required to prepare 300.0 mL of a 0.520 M solution.

0.300 L x 0.520 moles x 309.56 g = 48.3 g

1 L 1 mole

3. 150.0 g of NaCl is dissolved in 250.0 mL of water, calculate the molarity.

150.0 g x 1 mole

Molarity = 58.5 g (3sig figs) = 10.3 M

0.250 L

4. 25.2 g of CuSO₄^.6H₂O is dissolved in 28.0 mL of water, calculate the molarity.

25.2 g x 1 mole

Molarity = 267.72 g = 3.36 M

0.0280 L

5. Calculate the mass of NaCl required to prepare 565.0 mL of a 0.450 M solution.

0.5650 L x 0.450 moles x 58.5 g = 14.9 g

1 L 1 mole

6. Calculate the volume of 0.250 M NaCl solution required to contain

0.0300 g of NaCl.

0.0300 g NaCl x 1 mole x 1 L = 0.00205 L

58.5 g 0.250 mole

7. Calculate the volume of 0.500 M NaCl solution required to contain 0.52 g of NaCl.

0.52 g NaCl x 1 mole x 1 L = 0.018 L

58.5 g 0.500 mole

8. Calculate the mass of NaCl required to prepare 360.0 mL of a 0.35 M solution.

0.3600 L x 0.35 moles x 58.5 g = 7.4 g

1 L 1 mole

9. 55.6 g of NaCl is dissolved in 562 mL of water, calculate the molarity.

55.6 g x 1 mole

Molarity = 58.5 g = 1.69 M

0.562 L

10. 78.9 g of CuSO₄^.8H₂O is dissolved in 500.0 mL of water, calculate the molarity.

78.9 g x 1 mole

Molarity = 303.76 g = 0.519 M

0.5000 L

Stoichiometry Worksheet # 3

1. Excess sodium hydroxide solution is added to 20.0 mL of 0.184 M ZnCl₂, calculate the mass of zinc hydroxide that will precipitate.

NaOH_(aq) + ZnCl_2(aq) → Zn(OH)_2(s) + 2NaCl_(aq)

0.0200 L ? g

0.0200 L ZnCl₂ x 0.184 mole x 1 mole Zn(OH)₂ x 99.42 g = 0.366 g Zn(OH)₂

1 L 1 mole ZnCl₂ 1 mole

2. How many millilitres of 1.09 M HCl are required to react with a solution formed by dissolving 0.775 g of sodium carbonate?

Na₂CO_3(aq) + 2HCl_(aq) → 2NaCl_(aq) + H₂O_(l) + CO_2(g)

0.775g ? mL

0.775 g Na₂CO₃ x 1 mole x 2 mole HCl x 1 L x 1000 mL = 13.4 mL

106 g 1 mole Na₂CO₃ 1.09 moles 1 L

3. Calculate the number of grams barium carbonate that can be precipitated by adding 50.0 mL of 0.424 MBa(NO₃)₂.

Ba(NO₃)_2(aq) + K₂CrO_4(aq) → BaCrO_4(s) + 2KNO_3(aq)

0.0500 L Ba(NO)₂ x 0.424 mole x 1 mole BaCO₃ x 253.3 g = 5.37 g BaCrO₄

1 L 1 mole Ba(NO)₂ 1 mole

4. Determine the number of millilitres of 0.246 M AgNO₃ required to precipitate all the phosphate ion in a solution containing 2.10 g of sodium phosphate.

3AgNO_3(aq) + Na₃PO_4(aq) → Ag₃PO_4(s) + 3NaNO_3(aq)

156 mL

5. How many grams of silver nitrate must be used in the preparation of 150. mL of 0.125 M solution.

3.19 g

6. What volume of SO₂ is generated by the complete reaction of 35.0 mL of 0.924 M Na₂SO₃?

Na₂SO_3(aq) + 2NaOH_(aq) → 2NaCl_(aq) + H₂O_(l) + SO_2(g)

0.724 L

7. How many milliliters of 6.2 M NaOH must react to liberate 2.4 L of hydrogen at STP?

2Al_(s) + 6NaOH_(aq) → 2Na₃AlO_3(aq) + 3H_2(g)

35 mL

8. Calculate the weight of H₂C₂O₄^.2H₂O required to make 750.0 mL of a 0.480 M solution.

45.4 g

9. 25.4 L of HCl gas at STP are dissolved in 2.5 L of water to produce an acid solution. What volume of 0.200 M Ba (OH) ₂ will this solution neutralize?

2HCl + Ba(OH)₂ → BaCl₂ + 2HOH

25.4 L HCl x 1 mole x 1mole Ba (OH) ₂ x 1L = 2.8 L

22.4 L 2 mole HCl 0.200 mol

10. 8.25 L at STP of HCL gas is dissolved in 500 ml of water to produce an acid solution. What volume of 0.200 M Ca (OH) ₂ will this solution neutralize?

Ca (OH) ₂ + 2HCl → CaCl₂ + 2H₂O

8.25 L HCL x 1 mole x 1 mole Ca (OH) ₂ x 1 L = 0.921L

22.4 L 2 mole HCL 0.200 mol

11. 250 mL of water is added to 100 mL of 0.0200M H₂SO₄. What volume of 0.100M KOH will it neutralize?

H₂SO₄ + 2KOH → 2K₂SO₄ + 2HOH

0.1 L H2SO4 x 0.0200mol x 2 mole KOH x 1 L x 1000 mL

1L 1 mole H₂SO₄ 0.100 mol 1 L

= 40ml

12.     What volume of 0.924 M Na₂SO₃ is required for the production of 350.0 mL of SO₂ at STP?

          2HCl   +   Na₂SO_3(aq)       +  2NaOH_(aq)          →      2NaCl_(aq)    +          H₂O_(l) +     SO_2(g)

1.69 x 10^-2L

13.     How many milliliters of hydrogen at   STP  can be generated by 500.0 mL 6.2 M NaOH completely reacting with excess Al. 

          2Al_(s)  +        6NaOH_(aq)    →        2Na₃AlO_3(aq)            +         3H_2(g)

3.5  x 10⁴ mL

14.     64.5 L of HCl gas at STP are dissolved in water to produce an acid           solution. What volume of 0.200 M Ba (OH)₂ will this solution neutralize?

7.20 L

Titration Calculations Worksheet # 4

1. In a titration 12.5 mL of 0.200 M NaOH ia required to neutralize 10.0 mL of H₂SO₄.

          Calculate the concentration of the acid ?

H₂SO₄ + 2NaOH → Na₂SO₄ + 2HOH

          0.0100 L 0.0125 L

          ? M                             0.200 M         

    Molarity    =    0.0125 L NaOH x 0.200 moles x 1 mole H₂SO₄

   1 L   2 mole NaOH

                                                                                    0.0100 L

                     =        0.125 M         

2. In a titration 22.5 mL of 0.100 M HCl ia required to neutralize 20.0 mL of Ba(OH)₂ .

          Calculate the concentration of the base ?

2HCl + Ba(OH)₂ → BaCl₂ + 2HOH

          0.0225 L 0.0200 L

          0.100 M                           ? M         

    Molarity    =    0.0225 L HCl x 0.100 moles x 1 mole Ba(OH)₂

   1 L 2 mole HCl

                                                                                    0.0200 L

                     =        0.0563 M   

3. A burette filled with 1.52 M nitric acid solution reads 33.10 mL initially. After titrating a 25.00 mL sample of

          barium hydroxide the endpoint was reached and the burette showed 46.30 mL. What is the barium hydroxide

          concentration?

46.30 - 33.10 = 13.20 mL = 0.01320 L

          2HNO₃ +   Ba(OH)₂ → Ba(NO₃)₂ + 2HOH

          0.01320 L                0.02500 L

1.52 M    ? M

0.0132L HNO₃ x 1.52 mole   x 1 mole Ba(OH)₂

Molarity = 1 L   2 mole HNO₃ = 0.402 M

0.02500 L

4. A burette filled with 2.557 M sodium hydroxide solution reads 15.62 mL initially. After titrating a 25.00 mL sample of phosphoric acid the endpoint was reached and the burette now showed 39.22 mL. What is the [phosphoric acid]?

39.22 - 15.62 = 23.60 mL = 0.02360 L

3NaOH + H₃PO₄ → Na₃PO₄ + 3H₂O

          0 .02360 L 0.02500 L

2.557 M ? M

0.02360 L NaOH x 2.557 mole x 1 mole H₃PO₄

Molarity = 1 L 3 mole NaOH = 0.8046 M

0.02500 L

5. A 10.00 mL sample of 2.120 M sodium hydroxide solution is placed in a 250.0 mL Erlenmeyer flask. An indicator called bromothymol blue is added to the solution. The solution is blue. Hydrochloric acid was added from a burette until there was a green color (endpoint had been reached). Determine the concentration of hydrochloric acid given the following burette readings:

Burette final = 22.04 mL

Burette initial - 12.08 mL

Difference = 9.96 mL Beware subtraction!!!! One sig fig is lost!!!

NaOH + HCl → NaCl + H₂O

.01000 L .00996 L

2.120 M ? M

0.01000 L NaOH x 2.1220 mole x 1 mole HCl

Molarity = 1 L 1 mole NaOH = 2.13 M HCl

0.00996 L

6. The following data was obtained during the titration of 1.0097 M sodium hydroxide with a 25.00 mL aliquot of hydrofluoric acid:

Trial 1 Trial 2 Trial 3

Burette Final Reading 34.56 mL 39.42 mL 44.20 mL

Burette Initial Reading 14.94 mL 19.86 mL 24.66 mL

Vol. of NaOH Added

Use the above information to determine the concentration of the acid.

NaOH + HF → NaF + H₂O

          0 .01955 L                0 .0250 L

1.0097 M ? M

0.01955 L NaOH x 1.0097 mole x 1 mole HF

[HF] = 1 L 1 mole NaOH = 0.7896 M

0.0250 L

7. The following data was obtained during the titration of 0.0998 M sodium hydroxide with a

10.00 mL aliquot of sulphuric acid:

Trial 1 Trial 2 Trial 3

Burette Final Reading 26.05 mL 48.52 mL 33.78 mL

Burette Initial Reading 2.46 mL 34.94 mL 20.22 mL

Vol. of NaOH added

Use the above information to determine the concentration of the acid.

6.77 x 10^-2 M

8. The following data was obtained during the titration of 2.0554 M hydrochloric acid with a 25.00 mL aliquot of barium hydroxide:

Trial 1 Trial 2 Trial 3

Burette Final Reading   22.92 mL 25.32 mL 41.30 mL

Burette Initial Reading   0.06 mL 2.58 mL 18.54 mL

Volume of Acid Added

Use the above information to determine the concentration of the barium hydroxide.

0.9352 M

9.       What volume of 2.549 M NaOH is needed to fully titrate 50.0 mL of 1.285 M HCl solution ?

         HCl   +        NaOH                   →

         0.0500 L HCl  x   1.285 mol   x   1 mole NaOH      x   1L          =       0.0252 L

                                        1 L                1 mole HCl            2.549 mole

10.     What volume of 1.146 M KOH is needed to fully titrate 20.8 mL of 0.557 M H₂SO₄ solution ?

         H₂SO₄         +        2KOH                   →

         0.0208 L H₂SO₄   x   0.557 mol   x   2 mole KOH      x     1L      =        0.0202 L

                                            1 L                   1 mole H₂SO₄         1.146 mole

Titrations Worksheet # 5

1. Calculate the mass of H₂C₂O₄^.2H₂O required to prepare 500.0 mL of a 0.200M solution.

12.6 g

2. Calculate the mass of Cu₂SO₄^.6H₂O required to prepare 200.0 mL of a 0.300M solution.

19.9 g

3. In a titration 0.200 M NaOH is used to neutralize 10.0 mL of H₂SO₄. In three runs, the following data was collected. Calculate the concentration of the acid.

Volume of 0.20 M NaOH (mL)

Initial Burette Reading 12.90 15.70 18.50

Final Burette Reading 15.70 18.50 21.50

0.028 M

4. In a titration 0.250 M KOH is used to neutralize 25.0 mL of H₃PO₄. In three runs, the following data was collected. Calculate the concentration of the acid.

Volume of 0.250 M KOH (mL)

Initial Burette Reading 2.90 15.70 28.70

Final Burette Reading 15.70 28.70 42.70

12.80 13.00 14.00

Use 12.90 mL

0.0430 M

5. Calculate the volume of 0.500M H₃PO₄ required to neutralize 25.0 mL of

0.200M NaOH.

0.00333L

6. Calculate the volume of 0.50 M NaOH required to neutralize 35.0 mL of

0.100M H₂C₂O₄.

0.014L

7. In a titration 35.7 mL of 0.250 M H₃PO₄ is used to neutralize 25.0 mL of KOH. Calculate the molarity of the base.

2.08 M

8. In a titration 35.2 mL of 0.20 M H₂C₂O₄ is used to neutralize 10.0 mL of NaOH. Calculate the molarity of the base.

1.4 M

9. 2 Al + 3 I₂ → 2 AlI₃

Initial 12.0 mol 15.0 mol 0

Change:

End: 2.0 mol 0mol 10.0 mol

10. C + 2Cl₂ → CCl₄

Initial 16.0 mol 34.0 mol 0

Changes:

End: 0 mol 2.0 mol 16.0 mol

11. 4 Fe + 3 O₂ → 2 Fe₂O₃

Initial 12.0 mol 8.0 mol 0

Change:

End: 1.3 mol 0 mol 5.33 mol

12. 2 NO + O₂ → 2 NO₂

100. g x 1 mole 100. g x 1 mole 0

30.0 g 32.0 g

Init: 3.333 3.125 3.333

Change: 3.333 1.667 3.333

End: 0 1.458 mol x 32.0 g 3.333 x 46.0 g

1 mole 1 mole

Grams: 0 g 46.7 g 153 g

13. Calculate the volume of H₂ gas produced at STP by the reaction of 300. mL of 0.500 M HCl with excess Zn.

Zn + 2HCl → H₂ + ZnCl₂

0.300 L x 0.500 moles x 1 mole H₂ x 22.4 L = 1.68 L

1 L 2 mole HCl 1 mole

14. Calculate the volume of 0.30 M KCl solution that contains 9.00 g of KCl.

0.40 L

Dilutions Worksheet # 6

1. 20.0 mL of 0.200 M NaOH solution is diluted to a final volume of 100.0 mL, calculate the new concentration.

M₁V₁ = M₂V₂

(20.0)(0.200) = M₂(100.0)

M₂ = 0.0400 M

2. 15.0 mL of a solution of NaOH is diluted to a final volume of 250.0 mL and the new molarity is 0.0500 M. Calculate the original molarity of the base.

M₁V₁ = M₂V₂

(15.0) M₁ = (250.0) (0.0500)

M₁ = 0.833 M

3. 50.0 mL of 0.025 M NaOH solution is added to 150.0 mL of water. Calculate the new molarity.

V₂ = 50.0 mL + 150.0 mL = 200.0

M₁V₁ = M₂V₂

(0.025)(50.0) = M₂ (200.0)

M₂ = 0.0063M

4. 45.0 mL of a solution of NaOH is diluted by adding 250.0 mL of water to produce a new molarity of 0.0500 M. Calculate the molarity of the base.

0.328 M

5. A 0.125 M solution is concentrated by evaporation to a reduced final volume of 100.0 mL and a molarity of 0.150 M. Calculate the original volume.

120. mL

6. 850.0 mL of 0.280 M KOH solution is diluted to a final volume of 1000.0 mL, calculate the new concentration.

0.238 M

7. 95.0 mL of a solution of NaOH is diluted to a final volume of 135 mL and the new molarity is 0.0500 M. Calculate the original molarity of the base.

0.0711 M

Molarity Review # 7

1. Convert 250. g AgNO₃ to formula units and then to atoms of O.

2.66 x 10²⁴ at O

2. Convert 5.9 x10²⁵ H₂ molecules to grams.

2.0 x 10² g H₂

3. Calculate the percentage composition of MgSO₄.

20.2 % Mg 26.7 % S 53.2 % O

4. A compound is 42.3 % C, 5.94 % H, 32.9 % N, and18.8 % O and has a molecular mass of 425.25 g/mol. Calculate the empirical and molecular formula.

C₃H₅N₂O C₁₅H₂₅N₁₀O₅

5. How many grams O₂ are required to consume 56.3 g Al?

4Al + 3O₂ → 2Al₂O₃

50.0 g O₂

6. 25.5 mL of 0.100 M HCl reacts with excess Zn to produce 25.3 mL of H₂ gas at STP. Calculate the theoretical yield in mL and the percentage yield of H₂ gas.

Zn + 2HCl → H₂ + ZnCl₂.

0.0255 L x 0.100 mole x 1 mole H₂ x 22.4 L = 0.0286 L 88.6 %

1 L 2 mole HCl 1 mole

7. Calculate the energy produced by the complete reaction of 150. g H₂.

2H₂ + O₂ → 2H₂O + 130. KJ

4.83 x 10³ KJ

8. 84.0 g of Al reacts with 122 g O₂ to produce Al₂O₃. How many grams of Al₂O₃ are produced? Determine the mass of the reactant in excess and the limiting reactant.

4Al + 3O₂ → 2Al₂O₃

84.0 g x 1 mole 122 g x 1 mole

27 g 32 g

I 3.111 mole 3.8125 mole 0

C 3.111 mole 2.333 mole 1.5555 mole

E 0 1.4795 mole 1.5555 mole

47.3 g 159 g

9. 15.2 g of Al reacts with 14.3 g O₂ to produce Al₂O₃. How many grams of Al₂O₃ are produced? Determine the mass of the reactant in excess and the limiting reactant.

4Al + 3O₂ → 2Al₂O₃

15.2 g x 1 mole 14.3 g x 1 mole

27 g 32 g

I 0.5630 mole 0.4469 mole 0

C 0.5630 mole 0.4222 mole 0.2815 mole

E 0 0.247 mole 0.2815 mole

0.790 g 28.7 g

10. 15.8 g of KCl is dissolved in 225 mL of water. Calculate the molarity.

[KCl] = 15.8 g x 1 mole = 0.941 M

74.6 g

0.225 L

11. Calculate the mass of KCl required to prepare 250.0 mL of 0.250 M solution.

0.2500L x 0.250 mole x 74.6 g = 4.66 g KCl

1 L 1 mole

12. Calculate the volume of 0.30 M BaCl₂ solution that contains 6.00 g of KCl.

6.00 g x 1 mole x 1 L = 0.27 L

74.6 g 0.30 mole

13. Calculate the volume of 0.250 M H₃PO₄ that contains 0.250 g H₂SO₄.

0.250 g x 1 mole x 1 L = 0.0102 L

98.03 g 0.250 mole

14. 1.5 g of BaCl₂ is dissolved in 100.0 mL of water. Calculate the concentration.

0.072 M

15. How many moles of BaCl₂ are in 250.0 mL of a 0.200 M solution?

0.0500 moles

16. How many litres of a 0.200 MBaCl₂ solution contain 0.250 moles?

1.25 L

17. Calculate the volume of H₂ gas produced at STP by the reaction of 400.0 mL of 0.800 M HCl with excess Zn.

Zn + 2HCl → H₂ + ZnCl₂

3.58 L

18. Calculate the volume of 0.250 M H₃PO₄ required to neutralize 25.5 mL of 0.200 M NaOH.

H₃PO₄ + 3NaOH ® Na₃PO₄ + 3HOH

? L 0.0255 L

0.0255 L NaOH x 0.200 mole x 1 mole H₃PO₄ x 1 L = 0.00680 L

1 L 3 mole NaOH 0.250 mole

19. Calculate the volume of 0.500 M KOH required to neutralize 45.3 mL of 0.320 M H₂SO₄ .

H₂SO₄ + 2KOH ® K₂SO₄ + 2HOH

0.0453 L ? L

0.0453 L H₂SO₄ x 0.320 mole x 2 mole KOH x 1 L = 0.0580 L

1 L                         1 mole H₂SO₄ 0.500 mole

20. Calculate the mass of CoCl₃.6H₂O required to prepare 500.0 mL of a 0.200 M solution.

27.4 g

Worksheet # 8 Dilutions and Molarity

1. 40.0 mL of 0.400 M NaOH solution is diluted to a final volume of 200.0 mL, calculate the new concentration.

M₁V₁ = M₂V₂

(0.400)(40.0) = M₂(200.0)

M₂ = 0.0800 M

2. 85.0 mL of a solution of NaOH is diluted to a final volume of 290.0 mL and the new molarity is 0.0500 M. Calculate the original molarity of the base.

M₁V₁ = M₂V₂

M₁(85.0) = (0.0500)(290.0)

M₁ = 0.171 M

3. 150.0 mL of 0.025 M NaOH solution is added to 150.0 mL of water. Calculate the new molarity.

M₁V₁ = M₂V₂

(0.025)(150.0) = M₂(300.0)

M₂ = 0.013 M

4. 220.0 mL of a solution of NaOH is diluted by adding 250.0 mL of water to produce a new molarity of 0.0500 M. Calculate the molarity of the base.

M₁V₁ = M₂V₂

M₁(220.0) = (0.0500)(470.0)

M₁ = 0.107 M

5. A 0.350 M solution is concentrated by evaporation to a reduced final volume of 100.0 mL and a molarity of 0.825 M. Calculate the original volume.

V1 = 236 mL

6. 850.0 mL of 0.280 M KOH solution is diluted to a final volume of 1000.0 mL, calculate the new concentration.

M₂ = 0.238 M

7. 28 g of KCl is dissolved in 225 mL of water, calculate the molarity.

1.7 M

8. Calculate the mass of KCl required to prepare 125 mL of 0.450 M solution.

4.20 g

9. Calculate the volume of 0.40 M KCl solution that contains 8.00 g of KCl.

0.27 L

10. Calculate the volume of 0.400 M H₂SO₄ required to neutralize 25.0 mL of 0.200 M NaOH.

0.00625 L

11. Calculate the volume of H₂ gas produced at STP by the reaction of 250.0 mL of 0.600 M HCl with excess Zn. Zn + 2HCl → H₂ + ZnCl₂

1.68 L

12. 8.5 L of HCl gas at STP is dissolved in 325 mL of water, calculate the molarity of the acid solution.

1.2 M

13. How many moles of NaCl are in 350.0 mL of a 0.400 M solution?

0.140M

14. How many litres of a 0.300 M KCl solution contain 0.350 moles?

1.17 L

15. Calculate the mass of 8.25 x 10⁵ mL of H₂ gas at STP.

74.4 g

16. Calculate the number of formula units of KCl in 200.0 mL of 0.300 M solution.

3.61 x 10²² FU

Worksheet # 9   Ion Concentration

1. What is the concentration of each ion in a 10.5 M sodium sulphite solution?

Na₂SO₃ → 2 Na⁺ + SO₃^2-

10.5 M 21.0M 10.5M

2. What is the concentration of each ion in the solution formed when 94.5 g of nickel (III) sulphate is dissolved into 850.0 mL of water?

Ni₂ (SO₄) ₃ → 2Ni³⁺ + 3S0₄^2-

0.274 M 0.548M 0.822M

Molarity = 94.5 g x 1 mole

405.7 g = 0.274M

0.8500L

3. If 3.78 L of 0.960 M calcium fluoride solution is added to 6.36 L of water, what is the resulting concentration of each ion?

M₁V₁ = M₂V₂ CaF₂ → Ca²⁺ + 2F^-

(0.960) (3.78) = M₂ (10.14) 0.358 M 0.358 M 0.716 M

M₂= 0.358 M

4. What is the concentration of each ion in a 5.55 M zinc phosphate solution?

Zn₃ (PO₄) ₂ → 3Zn ²⁺ + 2PO₄^3-

5.55 M 16.7 11.1 M

5. What is the concentration of each ion in the solution formed when 94.78 g of iron (III) sulphate

is dissolved into 550.0 mL of water?

Fe₂ (SO₄)₃ → 2Fe³⁺ + 3SO₄^2-

0.4309 M 0.8619 M 1.293 M

94.78 g x 1 mol

[Fe₂ (SO₄) ₃] = 399.9 g = 0.4309 M

0.5500 L

6. If 6.25 L of 0.560 M sodium bromide solution is added to 3.45 L of water, what is the resulting

Concentration of each ion?

M₁V₁ = M₂V₂ NaBr → Na ⁺ + Br ^-

(0.560) (6.25) = M₂ (9.70) 0.361 M 0.361 M 0.361 M

M₂= 0.361 M

7. 50.0 mL of 0.200 M Na₃PO₄ solution is mixed with 150.0 mL of 0.400 M Na₂CO₃.

         Calculate all ion concentrations.

Na₃PO₄ → 3 Na⁺ + PO₄^3-

50.0          0.200 M     0.150 M 0.0500 M

200.0

        Na₂PO₄       → 2 Na⁺ + CO₃^2-

150.0       0.400 M      0.600 M 0.300 M

200.0

[Na⁺] = 0.600 M + 0.150 M = 0.750 M

8. What is the concentration of each ion in the solution formed when 16.5 g of Aluminum sulphate is

dissolved into 600.0 mL of water?

16.5 g x 1 mol

[Al₂ (SO₄) ₃] = 342.3 g =   0.0803 M

0.600 L

Al₂ (SO₄) ₃ → 2Al³⁺ + 3SO₄^2-

0.0803 M 0.161 M 0.241M

9. If 1.78 L of 0.420 M barium fluoride solution is added to 2.56 L of water, what is the resulting

concentration of each ion?

M₁V₁ = M₂V₂ BaF₂ → Ba²⁺ + 2F^-

(0.420)(1.78) = M₂ (4.34) 0.172 M 0.172 M 0.345 M

M_{2 =} 0.172 M

10. What is the concentration of each ion in a 1.22 M zinc acetate solution?

Zn (CH₃COO) ₂ → Zn ²⁺ + 2CH₃COO^-

1.22 M 1.22M 2.44M

11. What is the concentration of each ion in the solution formed when 94.78 g of cobalt (III) sulphate

is dissolved into 400.0 mL of water?

[Co₂(SO₄)₃] = 94.78 g x 1 mole

406.1 g = 0.5835 M

0.4000 L

Co₂(SO₄)₃ → 2Co³⁺ + 3SO₄^2-

0.5835 M 1.167 M 1.750 M

12. If the chloride concentration in 2.00 L of solution is 0.0900 M, calculate the [Al³⁺] and the molarity of the AlCl₃ solution.

AlCl₃ → Al³⁺ + 3Cl^-

0.0300 M 0.0300 M 0.0900 M

13. If the [Ga³⁺] concentration in 2.00 L of solution is 0.0300 M, calculate the [SO₄^2-] and the molarity of the Ga₂(SO₄)₃ solution.

Ga₂(SO₄)₃ → 2Ga³⁺ + 3SO₄^2-

0.0150 M 0.0300 M 0.0450 M

14. In a titration 12.5 mL of 0.200 M NaOH is needed to neutralize 10.0 mL of H₃PO₄, calculate the acid concentration.

H₃PO₄ + 3NaOH → Na₃PO₄ + 3HOH

0.0100 L 0.0125 L

? M 0.200 M

[H₃PO₄] = 0.0125 L NaOH x 0.200 mole x 1 moles H₃PO₄

1 L 3 mole NaOH = 0.0833 M

0.0100 L

15. What volume of 0.200 M H₂SO₄ is required to neutralize 25.0 mL of 0.300 M NaOH?

H₂SO₄ + 2NaOH → Na₂SO₄ + 2HOH

? L 0.0250 L

0.200 M 0.300 M

0.0250 L NaOH x 0.300 mole x 1 moles H₂SO₄ x 1 L = 0.0188 L

1 L 2 NaOH  0.200 mole

16. The [Cl^-] = 0.600 M in 100.0 mL of a AlCl₃ solution. How many grams AlCl₃ are in the solution?

AlCl₃ ® Al³⁺ + 3Cl^-

0.200 M 0.600 M

0.1000 L x 0.200 mole x 133.5 g = 2.67 g

1 L   1 mole

17. The [SO₄^2-] = 0.600 M in 100.0 mL of a Al₂(SO₄)₃ solution. How many grams Al₂(SO₄)₃ are in the solution?

6.85 g

Worksheet # 10 Molarity Unit Review # 1

1. 200.0 mL of 0.200 M H₂SO₄ reacts with 250.0 mL of 0.40 M NaOH, calculate the concentration of the excess base.

H₂SO₄ + 2NaOH → Na₂SO₄ + 2HOH

0.2000 L x 0.200 mole 0.250 L x 0.40 mole

1 L 1 L

I 0.0400 mole 0.100 mole

C 0.0400 mole 0.0800 mole

E 0 0.020 Note the loss of one sig fig!

[NaOH] = 0.020 mole = 0.044 M

0.4500 L Note that the final volume is 250.0 + 200.0 mL

2. 100.0 mL of 0.100 M H₂SO₄ reacts with 50.0 mL of 0.20 M NaOH, calculate the concentration of the excess acid.

H₂SO₄ + 2NaOH → Na₂SO₄ + 2HOH

0.1000 L x 0.100 mole 0.050 L x 0.20 mole

1 L 1 L

I 0.0100 mole 0.010 mole

C 0.0050 mole 0.010 mole

E 0.0050 mole 0

[H₂SO₄] = 0.0050 mole = 0.033 M

0.1500 L

Note that the final volume is 100.0 + 50.0 mL

3. 500.0 mL of 0.100 M H₂SO₄ reacts with 400.0 mL of 0.400 M NaOH, calculate the concentration of the excess base.

H₂SO₄ + 2NaOH → Na₂SO₄ + 2HOH

0.5000 L x 0.100 mole 0.4000 L x 0.40 mole

1 L 1 L

I 0.0500 mole 0.160 mole

C 0.0500 mole 0.100 mole

E 0 mole 0.060 mole

Note the loss of sig figs!

[NaOH] = 0.06 mole = 0.067 M

                                                            0.900 L

Note that the final volume is 500.0 + 400.0 mL

4. 100.0 mL of 0.200 M MgCl₂ reacts with 300.0 mL of 0.400 M AlCl₃, calculate all ion concentrations.

MgCl₂ → Mg²⁺ + 2Cl^-

100.0       0.200 M     0.0500 M 0.100 M

400.0

        AlCl₃       →    Al³⁺ +   3Cl^-

300.0       0.400 M      0.300 M 0.900 M

400.0

[Cl^-] = 0.100 M + 0.900 M = 1.000 M

5. Change 2.66 moles of H₂O to molecules.

1.60 x 10²⁴ molecules

6. Change 9.7x10¹⁹ atoms Fe to moles.

1.6 x 10^-4 mole

7. Convert 88.3 g AgNO₃ to formula units and then to atoms of O.

88.3 g x 1 mol x 6.02 x 10²³ FU x 3 atoms O = 9.39 x 10²³ atoms O

169.9 g 1 mol 1 FU

8. Convert 3.8 x 10²⁵ H₂ molecules to grams.

1.3 x 102 g

9. Calculate the empirical formula of a compound that is 62.2 % Pb, 8.454 % N, and 28.8 % O.

Is this compound ionic or covalent?

Pb (NO₃) ₂

10. A compound is 42.3 % C, 5.94 % H, 32.9 % N, and 18.8 % O and has a molecular mass of 425.25 g/mol. Calculate the empirical and molecular formula.

C₁₅H₂₅N₁₀O₅

11. How many moles of Al₂O₃ are produced by the reaction 200. g Al?

4Al + 30₂ → 2Al₂O₃

3.70 mole

12. How many moles Al are required to produce 300. g Al₂O₃?

4Al + 30₂ → 2Al₂O₃

5.88 mole

13. 100. g Al reacts with excess O₂ to produce 150. g Al₂O₃ according to

Calculate the theoretical and percentage yield. 4Al + 30₂ → 2 Al₂O₃.

79.4 %

14. Calculate the energy produced by the complete reaction of 150. g H₂.

2H₂ + O₂ → 2H₂O + 130. KJ

4.83 x 10³ kJ

15. How many grams of H₂ would be needed to produce 260. KJ of energy?

2H₂ + O₂ → 2H₂O + 130. KJ

8.08 g

16. 20. mol H₂ reacts with 8.0 mol O₂ to produce H₂O. Determine the number of grams reactant in excess and number of grams H₂O produced. Identify the limiting reactant.

8 g H₂ , 2.9 x 10² g H₂O

17. How many litres of O₂ gas are required to produce 100. g Al₂O₃?

4Al + 30₂ → 2Al₂O₃

32.9 L

18. 15.8 g of KCl is dissolved in 225 mL of water. Calculate the molarity.

0.941 M

19. Calculate the mass of KCl required to prepare 250.0 mL of 0.250 M solution.

4.66 g

20. Calculate the volume of 0.30 M KCl solution that contains 6.00 g of KCl.

0.27 L

21. Calculate the volume of 0.250 M H₂SO₄ required to neutralize 20.0 mL of 0.100 M NaOH.

0.00400 L

22. Calculate the volume of H₂ gas produced at STP by the reaction of 150.0 mL of 0.500 M HCl with excess Zn.

Zn + 2HCl → H₂ + ZnCl₂

0.840 L

23. 1.5 L of HCl gas at STP is dissolved in 225 mL of water, calculate the molarity of the acid solution.

0.30 M

24. How many moles of NaCl are in 250. mL of a 0.200 M solution?

0.0500 moles

25. How many litres of a 0.200 M KCl solution contain 0.250 moles?

1.25 L

26. Calculate the mass of 2.25 x 10⁵ mL of H₂ gas at STP.

20.3 g

27. Calculate the number of formula units of KCl in 100.0 mL of 0.200 M solution.

1.20 x 10²² FU

28. 40.6g of KBr is dissolved in 500.0 mL of water, calculate the molarity.

0.682 M

29. Calculate the mass of KBr required to prepare 450.0 mL of 0.350 M solution.

18.7 g

30. Calculate the volume of 0.50 M KCl solution that contains 3.00 g of KCl.

0.080 L

31. Calculate the volume of 0.250 M H₃PO₄ required to neutralize 25.5 mL of

0.200 M NaOH.

0.00680 L

32. In a titration 22.5 mL of 0.200 M H₃PO₄ is required to neutralize 10.0 mL of KOH. What is the molarity of the base?

1.35 M

Worksheet # 11 Molarity Unit Review # 2

1.         100.0 mL of 0.200 M HCl, 200.0 mL of 0.100 M HBr, and 175 mL of 0.100 M Ba(OH)₂.

             Calculate the concentration of the excess acid or base.

0.1000 L  x  0.200 mole   =    0.0200 mole HCl

                               1L

0.2000 L  x  0.100 mole   =    0.0200 mole HBr

                               1L

                                          =    0.0400 mole HX

0.175L  x  0.100 mole   =    0.0175 mole Ba(OH)₂

                               1L

            2HX              +           Ba(OH)₂ 

I           0.0400 mole                0.0175 mole

C        0.0350 mole           0.0175 mole

E         0.0050 mole 0.0000

Total Volume =  100.0 mL  +  200.0 mL  +  175 mL   =   475 mL

[HX] = 0.0050 mole

             0.475 L  

[HX] =  0.011 M 

2.      150.0 mL of 0.200 M HCl and 250 mL of 0.300 M HNO₃ react with excess CaCO₃. 

         Calculate the theoretical yield of CO₂. Start by writing an equation.

HCl         +      CaCO₃ ®      CO₂      +      CaCl₂          +    H₂O

0.1500 L HCl    x   0.200 mole    x    1 mole CO₂   x   44.0 g     =    0.660 g

                                     1 L                 2 mole HCl        1 mole

HNO₃        +      CaCO₃ ®      CO₂      +      Ca(NO₃)₂              +    H₂O

0.2500 L HNO₃    x   0.300 mole    x    1 mole CO₂   x   44.0 g     =    1.65 g

                                     1 L                    2 mole HCl        1 mole

Total               2.31 g

3. Calculate the percentage composition of Al₂(SO₄)₃ to three significant figures.

2 Al 54.0 %Al = 54.0 x 100% = 15.8 %

342.3

3 S 96.3 %S = 96.3 x 100% = 28.1 %

342.3

12 O 192.0 % O = 192.0 x 100% = 56.1%

342.3 342.3

4. A compound is 42.3 % C, 5.94 % H, 32.9 % N, and and18.8 % O and has a molecular mass of 850.5g/mol. Calculate the empirical and molecular formula.

42.3 g C x 1 mol = 3.525 mol = 3 C₃H₅N₂O → C₃₀H₅₀N₂₀O₁₀

12.0 g

5.94 g H x 1 mol = 5.881 mol = 5

1.01 g

32.9 g N x 1 mol = 2.350 mol = 2

14.0 g

18.8 g O x 1 mol = 1.175 mol = 1

16.0 g

5. How many grams of 0₂ are required to consume 56.3 g Al?

4Al + 30₂ → 2Al₂O₃

56.3 g Al x 1 mol x 3 mol O₂ x 32.0 g = 50.0 g O₂

27.0 g 4 mol Al 1 mol

6. 15.8 g of AlCl₃ is dissolved in 225 mL of water, calculate the molarity.

Molarity = 15.8 g x 1 mol

133.5 g = 0.526 M

0.225 L

7. Calculate the mass of AlCl₃ required to prepare 250.0 mL of 0.250 M solution.

0.250 L x 0.250 mol x 133.5 g = 8.34 g

L 1 mol

8. Calculate the volume of 0.30 M AlCl₃ solution that contains 6.00 g of AlCl₃.

6.00 g x 1 mol x 1 L = 0.15 L

33.5 g 0.30 mol

9. Calculate the volume of 0.450 M H₂SO₄ required to neutralize 25.0 mL of

0.200 M NaOH.

H₂SO₄ + 2NaOH → Na₂SO₄ + 2HOH

? L 0.025 L

0.025 L NaOH x 0.200 mol x 1 mol H₂SO₄ x L = 0.00556 L

L 2 mol NaOH 0.450 mol

10. Calculate the volume of H₂ gas produced at STP by the reaction of 350.0 mL of 0.600 M HCl with excess Zn.

Zn + 2HCl → H₂ + ZnCl₂

0.350 L HCl x 0.600 mol x 1 mol H₂ x 22.4 L = 2.35 L

L 2 mol HCl 1 mol

11. 2.9 L of HCl gas at STP is dissolved in 225 mL of water, calculate the molarity of the acid solution.

Molarity = 2.9 L x 1 mol = 0.58 M

22.4 L

0.225 L

12. How many moles of NaCl are in 500.0 mL of a 0.300 M solution?

0.500 L x 0.300 mol = 0.150 mol

L

13. How many litres of a .2300 M KCl solution contain 0.250 moles?

0.250 mol x L = 1.09 L

0.2300 mol

14. Calculate the mass of 560. mL of CO₂ gas at STP.

0.560 L x 1 mol x 44.0 g = 1.10g

22.4 L 1 mol

15. Calculate the number of formula units of NaCl in 100.0 mL of 0.200 M solution.

0.100 L x 0.200 mol x 6.02 x 10²³ FU = 1.20 x 10²² FU

L 1 mol

16. 25.5 mL of 0.100 M HCl reacts with excess Zn to produce 25.3 mL of H₂ gas at STP.

Calculate the theoretical yield in mL and the percentage yield of H₂ gas.

Zn + 2HCl → H₂ + ZnCl₂.

0.0255 L x 0.100 mol x 1 mol H₂ x 22.4 L = 0.0286 L

L 2 mol HCL 1 mol

% Yield = 25.3 x 100 % = 88.6 %

28.6

17. Calculate the energy produced by the complete reaction of 150. g H₂.

2H₂+O₂ → 2H₂O + 130KJ

150 g H₂ x 1mol x 130 KJ = 4.83 x 10³ KJ

2.02 g 2 mol

18. 84.0 g of Al reacts with 122g O₂ to produce Al₂O₃. How many grams of Al₂O₃ are produced? Determine the mass of the reactant in excess and the limiting reactant.

4 Al + 3O₂ → 2Al₂O₃

84.0 g x 1 mol 122g x 1mol

27.0 g 32.0 g

I 3.111 mol 3.813 mol 0

C 3.111 mol 2.333 mol 1.556

E 0 1.48 mol 1.556

Limiting 47.4 g O₂ excess                        159 g

19. Calculate the percentage composition of Na₂SO₄.

2 Na 46.0 g % Na = 32.4 %

1 S 32.1 g % S = 22.6 %

4 O 64.0 g % O = 45.0 %

142.1

20. How many litres of O₂ gas are required to produce 100. g Al₂O₃?

4Al + O₂ → 2Al₂O₃

100. g Al₂O₃ x 1 mole x 3 mole O₂ x 22.4 L = 32.9 L

102 g 2 mole Al₂O₃ 1 mole

21. Calculate the molar mass of a gas that weighs 19.43 g and has a STP volume of 9.894 L.

If the gas is a very funny one containing nitrogen and used by the dentist, determine the

molar mass and molecular formula for the gas.

9.894 L x 1 mole = 0.4417 mole

22.4 L

Molar Mass = 19.43 g = 44.0 g/mole N₂O

0.4417 mole

Write a balanced formula equation, complete ionic equation, and net ionic equation for each reaction. There are two no reactions.

22. Zn_{(s) +} 2AgNO_3(aq) → 2Ag_(s) + Zn(NO₃)_2(aq)

Zn_{(s) +} 2Ag⁺ + 2NO₃^- → 2Ag_(s) + Zn²⁺ + 2NO₃^-

Zn_{(s) +} 2Ag⁺ → 2Ag_(s) + Zn²⁺

23. BaS _(aq) + 2KOH_(aq) → Ba(OH)_2(s) + K₂S_(aq)

Ba²⁺ + S^2- + 2K⁺ + 2OH^- → Ba(OH)_2(s) + 2K⁺ + S^2-

Ba²⁺ + 2OH^- → Ba(OH)_2(s)

24. 2NaCl_(aq) + F_2(g) → 2NaF_(aq) + Cl_2(g)

2Na⁺ + 2Cl^- + F_2(g) → 2Na⁺ + 2F^- + Cl_2(g)

2Cl^- + F_2(g) → 2F^- + Cl_2(g)

25. Sr(OH)_{2 (aq)} + CuSO_4(aq) → Cu(OH)_{2 (s)} + SrSO_4(s)

Sr²⁺ + 2OH^- + Cu²⁺ + SO₄^2- → Cu(OH)_{2 (s)} + SrSO_4(s)

Sr²⁺ + 2OH^- + Cu²⁺ + SO₄^2- → Cu(OH)_{2 (s)} + SrSO_4(s)

26. NaCl_(aq) + Cu(NO₃)_2(aq) →

No reaction, both possible products have high solubility

27. NaCl(aq) + ZnF_2(aq) →

No reaction, both possible products have high solubility

28. 100.0 g of an aqueous compound that is 45.49 % Pb, 12.31 % N, and 42.20 % O reacts with another compound that is 28.16 % N, 8.13 % H, 20.79 % P, and 42.91 % O. If the actual yield of the product containing lead is 60.0 g, calculate the percentage yield.

60.0 g Actual Yield

3Pb(NO₃)_4(aq) + 4(NH₄)₃PO_4(aq) → Pb₃(PO₄)_4(s) + 12NH₄NO_3(aq)

100.0 g ? g Theoretical Yield

100.0 g Pb(NO₃)₄ x 1 mole x 1 mole Pb₃(PO₄)₄ x 1001.6 g = 75.3 g

455.2 g 3 mole 3Pb(NO₃)₄ 1 mole

% yield = 60.0 g x 100% = 81.8 %

29. The following data was obtained during the titration of 2.0554 M hydrochloric acid with a 25.00 mL aliquot of barium hydroxide:

Trial 1 Trial 2 Trial 3

Burette Final Reading 22.92 mL 25.32 mL 41.30 mL

Burette Initial Reading 0.06 mL 2.58 mL 18.54 mL

Vol. of Acid Added 22.86 mL 22.74 mL 22.76 mL

Use the above information to determine the concentration of the barium hydroxide.

2HCl + Ba(OH)₂ →

0.02275 L 0.02500 L

2.0554 M ? M

Molarity Ba(OH)₂ = 0.02275 L x 2.0554 mole x 1 mole Ba(OH)₂

1 L 2 mole HCl

0.02500 L

= 0.9352 M

30. 20.0 mL of 0.200 M NaOH solution is diluted to a final volume of 100.0 mL, calculate the new concentration.

M₁V₁ = M₂V₂

(0.200)(20.0) = M₂(100)

M₂ = 0.0400 M

31. 20.0 mL of 0.300 M AlCl₃ is mixed with 20.0 mL of 0.300 CaCl₂, calculate all ion concentrations.

ACl₃ → Al³⁺ + 3Cl^-

20.0         0.300 M     0.150 M 0.450 M

40.0

        CaF₂          → Ca²⁺ + 2Cl^-

20.0         0.300 M      0.150 M 0.300 M

40.0

[Cl^-] = 0.450 M + 0.300 M = 0.750 M

32. A burette filled with 2.000 M sodium hydroxide solution reads 20.20 mL initially. After titrating a

25.00 mL sample of phosphoric acid the endpoint was reached and the burette now showed 40.20 mL. What is the [phosphoric acid]?

H₃PO₄ + 3NaOH →

0.02500 L 0.02000 L

? M 2.000 M

Molarity H₃PO₄ = 0.02000 L x 2.000 mole x 1 mole H₃PO₄

1 L 3 mole NaOH

0.02500 L

= 0.5333 M

33. Calculate the volume of 0.500M KOH required to neutralize 45.0 mL of 0.320 M H₂SO₄.

H₂SO₄ + 2KOH →

0.0450 L ? L

0.0450 L H₂SO₄ x 0.320 moles x 2 mole KOH x 1 L = 0.0576 L

1 L 1 mole H₂SO₄ 0.500 mole

34. 100.0 mL of 0.200 M H₂SO₄ reacts with 150.0 mL of 0.40 M NaOH, calculate the concentration of the excess base.

H₂SO₄ + 2NaOH →

0.1000 L x 0.200 mole 0.1500 L x 0.40 mole

1 L 1L

I 0.0200 mole 0.060 mole

C 0.0200 mole 0.0400 mole

E 0 0.020 mole Note the loss of one sig fig

[NaOH] = 0.020 mole = 0.080 M

0.250 L

35. 200.0 mL of 0.10 M H₂SO₄ reacts with 100.0 mL of 0.20 M NaOH, calculate the concentration of the excess acid.

H₂SO₄ + 2NaOH →

0.2000 L x 0.100 mole 0.1000 L x 0.20 mole

1 L 1L

I 0.0200 mole 0.020 mole

C 0.010 mole 0.020 mole

E 0.010 loss of one sig fig 0

[H₂SO₄] = 0.010 mole = 0.033 M

0.300 L

36. 250.0 mL of 0.100 M H₂SO₄ reacts with 100.0 mL of 0.400 M NaOH and

            200.0 mL of 0.200 M KOH, calculate the concentration of the excess base.

H₂SO₄ + 2XOH →

0.2500 L x 0.100 mole 0.1000 L x 0.400 mole    +     0.2000 L x 0.200 mole

1 L 1L 1L

I 0.0250 mole 0.0800 mole

C 0.0250 mole 0.0500 mole

E 0 0.0300 mole

[NaOH] = 0.0300 mole = 0.0545 M

0.550 L

Worksheet # 12 Chemistry 11 Calculations Practice Test # 1

Pick two formulas that match each classification:

1. a b Acid a) HCl e) KOH

2. c d Covalent Nonacid b) CH₃COOH f) NH₄Cl

3. h f Salt c) CH₄ g) Ba(OH)₂

4. e g Base d) HOH h) AgNO₃

5. Calculate the molarity of the solution formed when 200 g of NaCl is dissolved in

100 mL of H₂O.

Molarity = 200g x 1 mole

58.5g = 34.2 M

0.100 L

6. How many grams of AgCl are required to prepare 150 mL of 0.200 M solution?

0.150L x 0.200 mole x 143.4 g = 4.30 g

           1 L 1 mole

7. How many litres of 0.200 M AgCl are needed to provide 50 g of AgCl?

50g x 1 mole x 1 L = 1.7 L

143.4g 0.200 mole

8. 100 g of AlCl₃ is dissolved in 200 mL H₂O, calculate [Al³⁺] and [Cl^-].

100 g x 1 mole

Molarity = 133.5 g = 3.745 M AlCl₃ → Al³⁺ + 3Cl^-

0.200 L

3.745 M 3.75 M 11.2M

9. In three runs of a titration 36.9, 34.4 and 34.3 mL of 0.200 M NaOH was required to neutralize a 25.0 mL sample of H₂CO₃. Calculate the molarity of the acid.

H₂SO₄ + 2NaOH → Na₂SO₄ + 2HOH

0.0250 L 0.3435 L

? M 0.200 M

[ H₂SO₄] = 0.03435 L x 0.200 mole x 1 mole H₂SO₄ = 0.137 M

1 L 2 mole NaOH

0.0250 L

= 0.137 M

10. Calculate the [NaOH] due to excess NaOH in the new solution produced by mixing 100. mL

0.200 M HCl and 100. mL 0.300 M NaOH.

HCL + NaOH → NaCl + HOH

0.100L x 0.200 mole = 0.0200 mol 0.100L x 0.300 mol = .030 mole

1 L 1 L

I 0.0200 mole 0.0300 mole

C 0.0200 mole 0.0200 mole

E 0 mole 0.0100 mole

Total Volume = 200 mL = 0.200 L Molarity = 0.0100 mole = 0.0500 M

0.200 L

11. A empty beaker has a mass of 29.86 g. The same beaker is filled with 0.250 L with a solution of CaCl₂ and weighs 87.26 g. The solution is evaporated to dryness and the mass of the beaker and solid is 62.31 g. Calculate the molarity of the solution.

Mass of CaCl₂ = 62.31 – 29.86 = 32.45g

Molarity = 32.45g x 1 mole

111.1g = 1.17 M

.250 L

12. Complete the reaction equations.

i) Formula Equation/Chemical Equation

2AgNO_{3 (aq) +} Na₂SO_{4 (aq)} → Ag₂SO_4(s) + 2NaNO_3(aq)

ii) Total Ionic Equation

2Ag⁺_(aq) + 2NO₃^- + 2Na⁺_(aq) + SO₄^2- → Ag₂SO_4(s) + 2Na⁺_(aq) + 2NO₃^-_(aq)

iii) Net Ionic Equation

2Ag⁺_(aq) + SO₄^2- → Ag₂SO_4(s)

13. Complete the formula equation:

2H₃PO_4(aq) + 3Sr(OH)_2(aq) → Sr₃(PO₄)_2(s) + 6HOH_(l)

Complete the complete ionic equation:

6H⁺_(aq) + 2PO₄^3- + 3Sr²⁺_(aq) + 6OH^- → Sr₃(PO₄)_2(s) + 6HOH_(l)

Complete the net ionic equation:

6H⁺_(aq) + 2PO₄^3- + 3Sr²⁺_(aq) + 6OH^- → Sr₃(PO₄)_2(s) + 6HOH_(l)

14. Complete the formula equation:

Fe₃(PO₄)_2(aq) + 3Zn_(s) → 3Fe_(s) + Zn₃(PO₄)_2(s)

Complete the complete ionic equation:

3Fe2+(aq) + 2PO43- + 3Zn(s) → 3Fe(s) + Zn3(PO4)2(s)

Complete the net ionic equation:

3Fe2+(aq) + 2PO43- + 3Zn(s) → 3Fe(s) + Zn3(PO4)2(s)

Worksheet # 13 Chemistry Calculations Practice Test # 2

1. Calculate the number of formula units in 250. g CaCl₂.

1.35 x 10²⁴ FU

2. Calculate the mass of 2.35 x 10²⁰ molecules of CO₂.

0.0172 g

3. Calculate the STP volume of 10.0 g of CO₂ gas.

5.09 L

4. Calculate the number of grams CaCl₂ in 350. mL of a 0.250M solution.

9.72g

5. Calculate the volume of 0.250 M NaCl solution that would contain 0.17 g NaCl.

0.012L

6. 1.26 g of AlCl₃ are dissolved in 160.0 ml of water. Calculate the molarity of the solution.

0.0590M

7. 12.5 ml of CO₂ gas at STP are dissolved in 250.0 ml of water. Calculate the molarity of the solution.

0.00223M

8. 10.0 g of Al₂(SO₄)₃ is dissolved in 155 ml of water. Calculate the two ion concentrations.

0.377 M

0.565 M

9. 200.0 ml of 0.200M H₃PO₄ reacts with 200.0 ml of 0.300M KOH. Calculate the molarity of the excess acid in the new solution formed.

0.0500M

10. 16 g of Ca react with water. Calculate the volume of H₂ gas produced at STP.

Ca + 2H₂O ® H₂ + Ca(OH)₂

8.9L

11. In a titration 0.200 M NaOH is used to neutralize 10.0 mL of H₂SO₄. In three runs the following data was collected. Calculate the concentration of the acid.

Volume of 0.200 M NaOH 25.3 mL 25.8 mL 25.6 mL

0.256 M

12. 60.0 g of Al react with 60.0 g of O₂. Calculate the amount of excess reactant.

4Al + 3O₂ → 2Al₂O₃

6.66 g O₂

13. Calculate the percentage composition of the elements in Ga₂ (SO₄)₃ to three significant digits.

32.6% , 22.5% , 44.9%

14. What volume of 0.300 M solution must be diluted to a final volume of 1200.0 mL and have a molarity of 0.2500M.

1.00L

15. Calculate the number of grams NaCl produced by the complete reaction of

520 g Cl₂. 2Na + Cl₂ → 2NaCl

857g

16. If the actual yield of NaCl in the last question was 200. g, calculate the percentage yield of NaCl.

23.3%

17. 200.0 ml 0.200 M HCl reacts with 400.0 ml 0.150M NaOH. Calculate the molarity of excess base.

HCl + NaOH → NaCl + H₂O

0.0333M

18. 100.0 mL of 0.250 M HCl solution is diluted by adding 250.0 mL of water, calculate the new concentration.

0.0714M

19. 65.5 mL of 0.300 M is diluted to a new molarity of 0.0600 M, how much water was added?

262mL

20. 56.0 mL of 0.100 M HCl reacts with 0.250 M Ba (OH) ₂, calculate the volume of base required to completely neutralize the acid.

0.0112L

21. Write the formula, complete, and net ionic equation for each.

H₃PO_{4 (aq)} and NaOH _(aq).

H₃PO_{4 (aq)} + 3NaOH _(aq) → Na₃PO_{4 (aq)} + 3HOH _(l)

3H⁺ _(aq) + PO₄^-3 _(aq) + 3Na⁺ _(aq) + 3OH^- _(aq) → 3Na⁺ _(aq) + PO₄^-3 _(aq) + 3HOH _(l)

H⁺ _(aq) + OH^- _(aq) → HOH _(l)

22. Write the formula, complete, and net ionic equation for each.

Na₃PO4 _(aq) and Ca (NO₃) _2(aq).

2Na₃PO_{4 (aq)} + 3Ca(NO₃) _{2 (aq)} → Ca₃ (PO₄) _{2 (s)} + 6NaNO_{3 (aq)}

Na⁺ _(aq) + 2PO₄^3- _(aq) +3Ca²⁺ _(aq) + 6NO₃^- _(aq) → Ca₃ (PO₄) _{2 (s)} + 6Na⁺ _(aq)+ 6NO₃^- _(aq)

3Ca²⁺ _(aq) + 2PO₄^3- _(aq) → Ca₃ (PO₄) _{2 (s)}

23. Write the formula, complete, and net ionic equation for each.

Cu (NO3) 2(aq) and Ag(s).

Ag _(s) + Cu (NO₃) _{2 (aq)} → No Reaction

24. A empty beaker has a mass of 25.86 g. The same beaker is filled with 0.250 L with a solution of Cl₂ and weighs 87.26 g. The solution is evaporated to dryness and the mass of the beaker and solid is 36.31 g. Calculate the molarity of the solution.

0.376 M

25. 125.0 g of an aqueous compound that is 3.091 % H, 31.62 % P, and 65.29 % O reacts with another compound that is 80.14 % Ba, 18.68 % O, and 1.179 % H. If the actual yield of the solid product is 350. g, calculate the percentage yield of the solid.

91.2%

26.        0.0250 M

27.        0.0091 M

28.       [Ca²⁺]   =    0.0714 M         [Al³⁺]  =  0.193 M          [Cl^-]  =    0.722M

  29        [Na⁺]   =    0.333 M         [PO₄^3-]  =  0.0667 M          [SO₄^2-]  =   0.0667 M

30.        0.0827 M

31.      150.0 mL of 0.200 M HCl and 250.0 mL of 0.300 M HNO₃ react with excess CaCO₃.

           Calculate the theoretical yield of CO₂. Start by writing an equation.

2.31 g