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?

H2C2O4              +            2KOH                 K2C2O4      +              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)  +  ZnSO4 (aq)    Zn(OH)2(s)     +     SrSO4(s)

ii) Total Ionic Equation

Sr2+    +    2OH-    +    Zn2+    +    SO42-         Zn(OH)2(s)     +     SrSO4(s)

iii) Net Ionic Equation

Sr2+    +    2OH-    +    Zn2+    +    SO42-         Zn(OH)2(s)     +     SrSO4(s)

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

Mg(s)   +   2HCl(aq)          H2(g)    +    MgCl2(aq)

Mg(s)      +   2H+      +       2Cl-          H2(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?

? 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   =    30.0 mL

1 L                       2 mole  KOH                 0.100 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                                           X2SO4      +              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