Chem. 12 Review Test 1 Answers

Pick two formulas that match each classification:

1.   e                   b                  Acid                                 a) KOH                  e) HCl

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

3.   h                   f                   Salt                                  c)  N2O                   g) Ba(OH)2

4.   a                   g                  Base                                 d)  HOH                 h)  AgNO3

5.       Calculate the molarity of the solution formed when 200.0 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 150.0 mL of 0.200 M solution?

.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 200.0 mL H2O, calculate [Al3+] and [Cl-].

100g   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

1 L               2 mole  NaOH

0.0250 L

=        0.137 M

10.     Calculate the molarity of the excess NaOH in the solution formed by mixing 50.0 mL of 0.200 M HCl with 50.0 mL of 0.200 M HNO3 and 75.0 mL of 0.300 M NaOH. Begin by writing a chemical equation for the reaction.

Assume that HCl will have the same effect on NaOH as HNO3 . Add the moles of HCl and HNO3 together.

HX            +                  NaOH   ®    NaX           +                  HOH

0.0500L HCl x  0.200 mole     =    0.0100 mol         0 .0750 L  x  0.300 mol  =   .0225 mole

1 L                                                                        1 L

0.0500L HNO3 x  0.200 mole  =   0.0100 mol

1 L

Total Acid

I                                    0.0200 mole                   0.0225 mole

C                                  0.0200 mole                    0.0200 mole

E                                   0  mole                           0.0025 mole     Note the loss of 1 sig fig

Total Volume = 175 mL = 0.175 L           Molarity     =         0.0025 mole           =  0.014 M

0.175 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

0.250 L

12.     Complete the reaction equations.

i) Formula Equation/Chemical Equation

2AgNO3 (aq)  +  Na2SO4 (aq)   ®      Ag2SO4(s)       +      2NaNO3(aq)

ii) Total Ionic Equation

2Ag+(aq)      +     2NO3-    +    2Na+(aq)     +  SO42-        ®     Ag2SO4(s)       +      2Na+(aq)       +      2NO3-(aq)

iii) Net Ionic Equation

2Ag+(aq)     +  SO42-        ®     Ag2SO4(s)

13.     Complete the formula equation:

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

Complete the complete ionic equation:

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

Complete the net ionic equation:

6H+(aq)      +     2PO43-     +    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)

15.     40.0 mL of 0.600 M AlCl3 is added to 80.0 mL of water. What are all ion concentrations?

AlCl3                                          ®                          Al3+              +          3Cl-

40.0  x  0.600 M   =        0.200 M                                    0.200 M                0.600 M

120.0

16.     200.0 mL of 0.150 M AlCl3 is added to 200.0 mL 0.250 M BaCl2, calculate the [Ba2+], [Al3+] and the [Cl-] immediately after mixing the two solutions.

AlCl3                                                    ®                          Al3+              +        3Cl-

200.0  x  0.150 M =        0.0750 M                                            0.0750 M                0.225 M

400.0

BaCl2                                        ®                          Ba2+             +          2Cl-

200.0  x  0.250 M =        0.125 M                                    0.125 M                0.250 M

400.0

[Ba2+]          =        0.125 M

[Al3+]           =        0.0750 M

[Cl-]             =        0.225 M      +        0.250 M      =        0.475 M

17.     A solution of 0.100 M H3C6H5O7(s) neutralized with 13.56 mL of 0.330 M KOH. Calculate the volume required.

H3C6H5O7   +        3KOH         ®      1K3C6H5O7 +        3HOH

?g                          0.01356 L

0.330 M

0.01356 L KOH    x   0.330 moles     x       1 mole H3C6H5O7     x      1 L           =        0.0149 L H3C6H5O7

1 L                3 mole KOH              0.100 mole