Question 4.10: Relating the Mass of a Product to the Volume and Molarity of......
Relating the Mass of a Product to the Volume and Molarity of a Reactant Solution
A 25.00 mL pipetful of 0.250 M K_{2}CrO_{4} is added to an excess of AgNO_{3}(aq). What mass of Ag_{2}CrO_{4} will precipitate from the solution?
K_{2}CrO_{4}(aq) + 2 AgNO_{3}(aq) → Ag_{2}CrO_{4}(s) + 2 KNO_{3}(aq)
Analyze
The fact that an excess of AgNO_{3}(aq) is used tells us that all of the K_{2}CrO_{4} in the 25.00 mL sample of K_{2}CrO_{4}(aq) is consumed. The calculation begins with a volume of 25.00 mL and ends with a mass of Ag_{2}CrO_{4} expressed in grams. The conversion pathway is mL soln → L soln → mol K_{2}CrO_{4} → mol Ag_{2}CrO_{4} → g Ag_{2}CrO_{4}.
Solve
Let’s solve this problem by using a stepwise approach.
Convert the volume of K_{2}CrO_{4}(aq) from milliliters to liters, and then use molarity as a conversion factor between volume of solution and moles of solute (as in Example 4-9). ? mol K_{2}CrO_{4} = 25.00 mL × \frac{1 L}{1000 mL} × \frac{0.250 mol K_{2}CrO_{4}}{1 L}
= 6.25 × 10^{−3} mol K_{2}CrO_{4}
Use a stoichiometric factor from the equation to convert from moles of K_{2}CrO_{4} to moles of Ag_{2}CrO_{4}. ? mol Ag_{2}CrO_{4} = 6.25 × 10^{−3} mol K_{2}CrO_{4} × \frac{1 mol Ag_{2}CrO_{4}}{1 mol K_{2}CrO_{4}}
= 6.25 × 10^{−3} mol Ag_{2}CrO_{4}
Use the molar mass to convert from moles to grams of Ag_{2}CrO_{4}. ? g Ag_{2}CrO_{4} = 6.25 × 10^{−3} mol Ag_{2}CrO_{4} × \frac{331.7 g Ag_{2}CrO_{4}}{1 mol Ag_{2}CrO_{4}}
= 2.07 g Ag_{2}CrO_{4}
The same final answer can be obtained more directly by combining the steps into a single line calculation.
? g Ag_{2}CrO_{4} = 25.00 mL × \frac{1 L}{1000 mL} × \frac{0.250 mol K_{2}CrO_{4}}{1 L} × \frac{1 mol Ag_{2}CrO_{4}}{1 mol K_{2}CrO_{4}} × \frac{331.7 g Ag_{2}CrO_{4}}{1 mol Ag_{2}CrO_{4}}
= 2.07 g Ag_{2}CrO_{4}
Assess
The units work out properly, which is always a good sign. As we have done before, let’s work the problem in reverse and use numbers that are rounded off slightly. A 2 g sample of Ag_{2}CrO_{4} contains 2/332 = 0.006 moles of Ag_{2}CrO_{4}. The number of moles of K_{2}CrO_{4} in the 25 mL sample of the K_{2}CrO_{4} solution is also approximately 0.006 moles; thus, the molarity of the K_{2}CrO_{4} solution is approximately 0.006 mol/0.025 L = 0.24 M. This result is close to the true molarity (0.250 M), and so we can be confident that our answer for the mass of Ag_{2}CrO_{4} is correct.