Chem. 12 Review Test #2 Name:

Block:

Pick two formulas that match each classification:

1. b g Acid a) Ba(NO₃)₂ e) NaBr

2. c f Covalent Nonacid b) CH₃COO H f) C₃H₈

3. a e Salt c) H₂O g) H₃PO₄

4. d h Base d) Al(OH)₃ h) Ca(OH)₂

5. A student wants to determine the molarity of 100.0 mL of NaCl solution. She weighs another empty beaker and determines the mass to be 56.23g. She transfers all of the solution to this beaker and weighs it and finds the mass to be 159.99g. She proceeds to evaporate off the water until it is dry and measures the mass of the beaker and residue to be 58.69g. What is the molarity of the solution?

58.69 – 56.23 = 2.46

Molarity = 2.46 g x 1 mole

58.5 g = 0.421 M

.100 L

6. How many grams of AlCl₃ are required to prepare 250 mL of 0.200 M solution?

0.250L x 0.200 mole x 133.5 g = 6.68 g

1 L 1 mole

7. How many litres of 0.350 M MgCl₂ are needed to provide 250.0 g of MgCl₂?

250 g x 1 mole x 1 L = 7.50 L

95.3 g 0.350 mole

8. 50.6 g of AlCl₃ is dissolved in 250.0 mL H₂O, calculate [Al³⁺] and [Cl^-].

50.6 g x 1 mole

Molarity = 133.5 g = 1.516 M AlCl₃ ® Al³⁺ + 3Cl^-

0.250 L 1.516 M 1.52 M 4.55 M

9. 600 mL of 0.200 M H₂SO₄ reacts with 600 mL of 0.200 M NaOH. Calculate concentration of the excess acid in the new solution.

H₂SO₄ + 2NaOH ® Na₂SO₄ + 2HOH

0.600L x 0.200 mole = .120 mol 0.600L x 0.200 mol = .120 mole

1 L 1 L

I 0.120 mole 0.120 mole

C - 0.060 mole - 0.120 mole

E = 0.060 mole = 0.000 mole

Beware subtraction- note the loss of 1 significant figure!

Sometimes numbers disappear during subtraction!

Total Volume = 1200 mL = 1.20 L Molarity = 0.060 mole = 0.050 M

1.20 L

10. In three runs of a titration 0.200 M NaOH was used to neutralize a 25.0 mL sample of H₂CO₃. Calculate the molarity of the acid.

0.200 M NaOH in the burette
Initial burette reading (mL)	2.05	10.56	19.09
Final burette reading (mL)	10.56	19.09	27.80
	8.51	8.53	8.71 reject

H₂CO₃ + 2NaOH → Na₂SO₄ + 2HOH

0.0250 L 0.0.00852 L

? M 0.200 M

[ H₂SO₄] = 0.00852 L NaOH x 0.200 mole x 1 mole H₂CO₃

1 L 2 mole NaOH

0.0250 L

= 0.0341 M

11. How many grams of .0200M H₂C₂O₄ are required to neutralize 250 mL of .0250M KOH?

H₂CO₃ + 2KOH → K₂SO₄ + 2HOH

0.250L KOH x 0.0250 mole x 1 mole H₂C₂O₄ x 90.0 g = 0.281 g

1 L 2 mole KOH 1 mole

12.Complete the reaction equations.

i) Formula Equation/Chemical Equation

Sr(OH)_{2 (aq)} + ZnSO_{4 (aq)} → Zn(OH)_2(s) + SrSO_4(s)

ii) Total Ionic Equation

Sr²⁺ + 2OH^- + Zn²⁺ + SO₄^2- → Zn(OH)_2(s) + SrSO_4(s)

iii) Net Ionic Equation

Sr²⁺ + 2OH^- + Zn²⁺ + SO₄^2- → Zn(OH)_2(s) + SrSO_4(s)

13. Write the complete ionic equation for the reaction of Mg _(s) and HCl _(aq).

Mg_(s) + 2HCl_(aq) → H_2(g) + MgCl_2(aq)

Mg_(s) + 2H⁺ + 2Cl^- → H_2(g) + Mg²⁺ + 2Cl^-

14. What volume of 0.100 M H₂SO₄ is needed to neutralize 25.0 mL 0.250 M NaOH and 30.0 mL of 0.200 M KOH solution?

Start with NaOH H₂SO₄ + 2NaOH → Na₂SO₄ + 2HOH

? L 0.0250 L

0.100 M 0.250 M

= 0.0250 L NaOH x 0.250 mole x 1 mole H₂SO₄ x 1L x 1000 mL = 31.3 mL

1 L 2 mole NaOH 0.100 mole 1 L

Now do KOH H₂SO₄ + 2KOH → Na₂SO₄ + 2HOH

? L 0.0250 L

0.100 M 0.250 M

= 0.0300 L KOH x 0.200 mole x 1 mole H₂SO₄ x 1L x 1000 mL = 15.0 mL

1 L 2 mole KOH 0.200 mole 1 L

Total Volume of H₂SO₄ = 31.3 + 15.0 = 46.3 mL

15. Calculate all ion concentrations after 200.0 mL of 0.200 M CaCl₂ is mixed with 200.0 mL of 0.300 M AlCl₃.

CaCl₂ ® Ca²⁺ + 2Cl^-

200.0 x 0.200 M = 0.100 M 0.200 M

400.0

AlCl₃ ® Al³⁺ + 3Cl^-

200.0 x 0.300 M = 0.150 M 0.450 M

400.0

[Ca²⁺] = 0.100 M

[Al³⁺] = 0.150 M

[Cl^-] = 0.200 M + 0.450 M = 0.650 M

16. What concentration of acid or base remains when 200.0 mL of 0.200 M H₂SO₄ is mixed with 200.0 mL 0.100 M KOH and 400.0 mL 0.100 M NaOH.

H₂SO₄ + 2XOH → Na₂SO₄ + 2HOH

0.200 L x 0.200 mol = 0.0400 moles 0.200 L x 0.200 mol = 0.0200 moles Add NaOH and KOH

1 L 1 L

0.400 L x 0.100 mol = 0.0400 moles

1 L

I 0.0400 mole 0.0600 mole

C 0.0300 mole 0.0600 mole

E 0.0100 mole 0.000 mole

Final Volume = 200.0 mL + 200.0 ml + 400.0 mL = 800.0 mL

[H₂SO₄] = 0.0100 mole = 0.0125 M

0.800 L