Review of Chem 11

 

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In order to be successful in Chemistry 12, there are several key concepts from Chemistry 11 that you must master. We will spend two periods in class reviewing these concepts. I challenge you to get 86 % or higher on this test. I believe that everyone in this class can get this mark. You need to convince yourself that you can, then do then work and you will. This worksheet contains everything that will be tested. I can give you a practice test if you want one. If you need help, that’s what I am here for- see me.

 

HCl     C3H8        SO2    NH4Cl    KOH    H2SO4    H2O    AgNO3    PbSO4     H3PO4    Ca(OH)2     Al(OH)3    P2O5    Ba(OH)2   CH3COOH

 

1. Classify the above as ionic or covalent by making two lists. Describe the difference between an ionic and covalent compound.

 

Ionic                            NH4Cl    KOH    AgNO3    PbSO4     Ca(OH)2     Al(OH)3    Ba(OH)2   

 

Covalent                     C3H8       SO2        H2O             P2O5      H3PO4 CH3COOH     H2SO4

 

1. Classify the above as acids, bases, salts and molecular (covalent compounds) by making four lists.

 

Acids                           HCl                 H2SO4                                      CH3COOH                  H3PO4

 

Bases                          KOH               Ca(OH)2                     Ba(OH)2                      Al(OH)3   

 

Salts                            NH4Cl             AgNO3                        PbSO4                            NaCl       

 

Molecular                   C3H8                SO2                            H2O                             P2O5

 

 

2. Calculate the molar mass of FeSO4 • 5H2O and Co3(PO4)2 • 6H2O.

 

241.9 g/mol           474.7 g/mol 

 

3. 0.300 moles of NaCl is dissolved in 250.0 ml of water, calculate the molarity.

 

 

Molarity =    0.300 moles          

                                               =      1.20 M    

                      .250 L

 

 

4. 500. g of FeSO4.6H2O is dissolved in 600. ml of water, calculate the molarity.

 

 

Molarity = 500g   x   1 mole

                                  259.9g           =     3.21 M

                

                                   .600 L

 

 

5. How many grams of NaCl are required to prepare 100.0 ml of a 0.200 M solution?

 

                .100L   x   0.200 mole   x    58.5 g       =           1.17g  

                                                 1 L              1 mole

 

 

6. 20. g of MgCl2 are dissolved in 250. ml of water, calculate the concentration of each ion.

 

                  20g   x  1 mole

Molarity =                     95.3 g           =     0.84 M                   MgCl2                →          Mg2+                 +       2Cl-

                                    

                                  0.250 L                                                      0.84 M                            0.84 M                  1.7 M  

 

 

 

7. How many liters of 0.300 M NaCl contains 10.0 g of NaCl?

 

10.0g   x      1 mole        x         1 L                 =    0.570 L

                              58.5g               0.300 mole

 

 

8. For each double replacement reaction write the formula equation, the complete ionic equation and the net ionic equation.

 

a)         2H3PO4(aq)     +          3Sr(OH)2(aq)               →      Sr3(PO4)2(s)      +     6HOH(l)

 

            6H+(aq)          +     2PO43-(aq)          +   3Sr2+(aq)          +     6OH-(aq)     →    Sr3(PO4)2(s)      +     6HOH(l)

 

                    6H+(aq)          +     2PO43-(aq)          +   3Sr2+(aq)          +     6OH-(aq)     →   Sr3(PO4)2(s)      +     6HOH(l)

 

b)         3Ca(NO3)2(aq)             +          2Na3PO4(aq)                →      Ca3(PO4)2(s)     +     6NaNO3(aq)

 

            3Ca2+(aq)    +   6NO3-(aq)    +   6Na+(aq)     +    2PO43-(aq)   →  Ca3(PO4)2(aq)     +     6Na+(aq)     +    6NO3-(aq)   

 

            3Ca2+(aq)    +   2PO43-(aq)   →  Ca3(PO4)2(aq)    

 

c)         Zn(s)      +      2HCl(aq)   →  H2(g)     +    ZnCl2(aq)

 

            Zn(s)      +      2H+(aq)    +   2Cl-(aq)     →  H2(g)     +    Zn2+(aq)     +     2Cl-(aq)    

 

            Zn(s)      +      2H+(aq)   →   H2(g)     +    Zn2+(aq)    

 

 

9. In three runs of a titration 22.8, 22.1 and 22.2 ml of .200 M Ba(OH)2 were required to neutralize 10.0 ml of HCl, calculate the acid concentration.

 

                                               2HCl              +               Ba(OH)2    →         BaCl2      +              2HOH

 

                                               0.0100 L                          0.02215 L

                                               ? M                                 0.200 M

 

                                        [HCl]           =        0.02215 L   x   0.200 mole   x  2 mole HCl         

                                                                                                           1 L              1 mole Ba(OH)2   

 

                                                                                                                     0.0100 L

                                                                  

=          0.886 M 

 

 

 

 

 

 

10. In three runs of a titration 12.1, 12.8, 12.8 ml of 0.200 M HCl  were required to neutralize 10.0 ml of Ca(OH)2, calculate the base concentration.

 

                                               2HCl              +    Ca(OH)2    →         CaCl2  +          2HOH

 

                                               0.0128                             0.02215 L

                                               0.200M

 

                                        Molarity      =        0.0128 L HCl   x   0.200 mole   x       1 mole Ca(OH)2   

                                                                                                           1 L                       2 mole HCl         

 

                                                                                                                     0.0100 L

                                                                  

=          0.128 M 

 

11. 35.0 ml of 1.00 M H2SO4 reacts with 175 ml  0.250M NaOH, calculate the concentration of the excess base.

 

                                     H2SO4              +              2NaOH    →         Na2SO4      +              2HOH

 

0.0350L x 1.00 mole  =  0.0350 mole         0.175L x 0.250 mole  =   0.04375 mole

                     1 L                                                             1 L

 

I                                    0.0350 mole                    0.04375 mole

 

C                                  0.02188 mole                  0.04375 mole

 

E                                   0.01312 mole                  0 mole

 

Total Volume = 210 mL = 0.210 L           Molarity     =              0.01312 mole           =  0.0625 M

                                        0.210L

 

 

 

 

12. 350.0 ml of 0.200 M HCl reacts with 175 ml  0.125 M Ca(OH)2, calculate the concentration of the excess acid.

 

 

                                     2HCl              +              Ca(OH)2    →         CaCl2      +              2HOH

 

0.350L x 0.200 mole  =  0.0700 mole         0.175L x  0.125 mole  =   0.021875 mole

                     1 L                                                             1 L

 

I                                    0.0700 mole         0.021875 mole

 

C                                  0.04375 mole         0.021875 mole

 

E                                   0.02625 mole        0.000 mole

 

Total Volume = 525 mL = 0.525 L           Molarity     =         0.02625 mole           =  0.0500 M

                                                                                                    0.525L

 

 

 

13. 25.0 g of sodium reacts with water, how many grams of hydrogen are produced?

 

                        2Na(s)           +   2HOH(l)  →    H2(g)     +    2NaOH(aq)     

 

 

 

25.0 g Na  x      1 mole      x    1mol H2    x        2.02 g       =    1.10 g

                                                                            23.0 g            2mol Na             1 mole

 

 

 

14. 25.0 g of calcium reacts with water, how many grams of calcium hydroxide are produced?

 

                        Ca(s)  +   2HOH(l)  →   H2(g)     +    Ca(OH)2(aq)     

 

 

 

25.0 g Ca  x      1 mole      x    1mol Ca(OH)2    x        74.1 g       =    46.2 g

                                                                             40.1 g            1mol Ca                      1 mole

 

 

 

 

 

 

15. How many millilitres of 0.200M NaOH is required to neutralize 25.0 ml of  0.100 M H2SO4 ?

 

 

                 H2SO4              +             2NaOH    →         Na2SO4      +              2HOH

 

                 0.025 L                            ? L

                 0.100 M                          0.200 M

 

 

 

0.0250 L H2SO4      x   0.100 mole   x   2 mole  NaOH           x   1 L             x    1000 mL       =    25.0 mL

                                         1 L                1 mole H2SO4                 0.200mole            1L

 

 

 

 

16. How many millilitres of 0.200M H2SO4 is required to neutralize 25.0 ml of 0.100 M NaOH ?

 

 

                 H2SO4              +             2NaOH    →         Na2SO4      +              2HOH

 

                 ? mL                               0.0250 L

                 0.200 M                          0.100 M

 

 

 

0.0250 L NaOH      x      0.100 mole   x       1 mole H2SO4    x      1 L               x     1000 mL       = 6.25 mL

                                      1 L                       2 mole  NaOH           0.200mole               1L

 

 

 

17. If the [F-] = 0.600 M in a AlF3  solution, calculate the [Al+3] and the number of grams required to make 1.00 L of the solution.

 

AlF3          →         Al3+       +       3F-

 

0.200 M            0.200 M           0.600M

               

1.00L   x   0.200 mole   x    84.0 g       =    16.8 g

                                  1 L              1 mole

 

 

 

18.  If the [Na+] = 0.250 M in a Na3P  solution, calculate the [P-3] and the number of grams required to make 1.50 L of the solution.

 

Na3P       →    3Na+            +           P3-

                     

                      0.250 M              0.08333M

               

1.50L   x   0.08333 mole   x      100 g       =    12.5 g

                                  1 L                   1 mole

 

 

 

 

 

19. A beaker of mass = 25.36g contains 2.00 L of a solution of BaCl2 and is evaporated to dryness mass = 28.59 g. Calculate the molarity of the solution.

 

28.59g – 25.36 =3.23g

 

Molarity = 3.23g   x   1 mole

                                   208.3g           =     0.00775 M

                

                                   2.00 L

 

 

20.       200.0 mL of 0.100 M Na3PO4 is mixed with 100.0 mL of 0.200 M Na2SO4. Calculate all ion concentrations. There is no reaction in the situation as the solutions are just diluting each other and the [Na+] increases.  First write out two dissociation equations. Use dilution factors to lower each concentration. Add the two [Na+] concentrations together at the end to get the total.

 

 

                    Na3PO4                 →                3Na+            +                  PO43-

 

            200    0.100 M  =  0.0667 M              0.200 M                          0.0667 M

            300

Na2SO4                 →                2Na+            +                  SO42-

 

            100    0.200 M  =  0.0667 M              0.133 M                          0.0667 M

            300

 

                    [Na+]  =  0.200 M  +  0.133 M  =  0.333 M    Ding Ding Ding!

 

21. A titration was performed by adding 0.175 M H2C2O4 to a 25.00 mL sample of NaOH. The following data was collected. Calculate the molarity of the base.

 

Trial #1

Trail # 2

Trial #3

Final volume of H2C2O4 (mL)

23.00

39.05

20.95

Initial volume of H2C2O4 (mL)

4.85

23.00

5.00

                                                                                                            18.15 mL                    16.05 mL                    15.95 mL       Average 16.00 mL

 

 

                                               H2C2O4              +           2NaOH    →         Na2C2O4      +              2HOH

 

                                               0.0160 L                          0.0250 L

                                               0.175 M                                  ?M

 

                              [ NaOH]                =        0.0160 L   x   0.175 mole   x    2 mole  NaOH

                                                                                                         1 L                1 mole H2C2O4      

 

                                                                                                                     0.0250 L

                                                                  

=        0.224 M

 

 

 

 

 

 

22.  2HCl     +    Ba(OH)2   →   BaCl2    +    2 H2O

When 3.16 g samples of Ba(OH)2 were titrated to the endpoint with HCl solution. 37.80mL, 35.49mL, 35.51 mL of HCL was required. Calculate the HCl concentration.

 

2HCl         +        Ba(OH)2  →         BaCl2          +        2 H2O

 

0.03550 L           3.16g

? M                    171.3g/mole

 

                              [HCl]           =        3.16g     x   1 mole   x     2 mole HCl

                                                                                       171.3 g        1 Ba(OH)2  

 

                                                                             0.03550 L

                                                                  

=        1.04 M