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