Final Exam Review Problem #1 KEY

Solve the following problem. In this problem, look at the chemistry skills that you will have to employ:

50.0 mL of a 0.300 M aqueous solution of sodium sulfate reacts with 30.0 mL of a 0.100 M aqueous solution of barium nitrate.

(a) What is the mass (in grams) of the barium compound that precipitates out of solution?

(b) What additional reactant and how much more would be required to produce the maximum amount of barium compound precipitate?

1 Na2SO4 (aq) + 1 Ba(NO3) (aq) --> 1 BaSO4 + 2 NaNO3
50.0 mL             30.0 mL? g
0.300 M             0.100 M

50.0 mL    x     1 L      x      .300 mole Na2SO4    x   1 mole BaSO4     x   233.39 g BaSO4
                      103 mL             1 L                              1 mole Na2SO4             1 mole BaSO4

= 3.50085 g BaSO4 --> 3.50 g (3 SF)

30.0 mL    x     1 L     x    .100 mole Ba(NO3)2      x  1 mole BaSO4   x     233.39 g BaSO4
                       103 mL             1 L                               1 mole Ba(NO3)2         1 mole BaSO4

= .70017 g BaSO4 --> .700 g (3 SF)

(a) Therefore, the 30.0 mL of the barium nitrate solution is the limiting reactant and so 0.700 g of the barium sulfate (1 SF) precipitates out of solution.

(b) Since the barium nitrate is the limiting reactant, you'd need more of this in order to use up the rest of the sodium sulfate. So, we'll work backwards now to determine how much more barium nitrate we'll need to use up all the sodium sulfate.

3.50 g BaSO4    x     1 mole BaSO4    x     1 mole Ba(NO3)2    x     1 L      x 103 mL
                                   233.39 g BaSO4           1 mole BaSO4          .1 mole        1 L
= 149.96 mL needed so 150. mL (3 SF) of a .100 M solution


So, since you started out with 30.0 mL and you need 150. mL you'll need 120. mL more of the barium nitrate solution!!

Another way to attack this problem is to take care of the subtraction FIRST:

3.50 g - .700g = 2.80 g MORE

Then, starting with 2.80 g of barium sulfate, calculate that volume:

2.80 g BaSO4 x 1 mole BaSO4 x 1 mole Ba(NO3)2 x 1 L x 103 mL =
233.39 g BaSO4 1 mole BaSO4 .1 mole 1 L

120. mL more of the barium nitrate solution

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