Question 4.11: Determining the Limiting Reactant in a Reaction. Phosphorus ......
Determining the Limiting Reactant in a Reaction
Phosphorus trichloride, PCl_{3}, is a commercially important compound used in the manufacture of pesticides, gasoline additives, and a number of other products. A ball-and-stick model of PCl_{3} is shown below. Liquid PCl_{3} is made by the direct combination of phosphorus and chlorine.
P_{4}(s) + 6 Cl_{2}(g) → 4 PCl_{3}(l)
What is the maximum mass of PCl_{3} that can be obtained from 125 g P_{4} and 323 g Cl_{2}?
Analyze
The following conversions are required: mol limiting reactant → mol PCl_{3} → g PCl_{3}. The key to solving this problem is to correctly identify the limiting reactant. One approach to this problem, outlined in Figure 4-9, is to compare the initial mole ratio of the two reactants with the ratio in which the reactants combine—6 mol Cl_{2} to 1 mol P_{4}. If more than 6 mol Cl_{2} is available per mole of P_{4}, chlorine is in excess and P_{4} is the limiting reactant. If fewer than 6 mol Cl_{2} is available per mole of P_{4}, chlorine is the limiting reactant.
Solve
? mol Cl_{2} = 323 g Cl_{2} × \frac{1 mol Cl_{2}}{70.91 g Cl_{2}} = 4.56 mol Cl_{2}
? mol P_{4} = 125 g P_{4} × \frac{1 mol P_{4}}{123.9 g P_{4}} = 1.01 mol P_{4}
Since there is less than 6 mol Cl_{2} per mole of P_{4}, chlorine is the limiting reactant. The remainder of the calculation is to determine the mass of PCl_{3} formed in the reaction of 323 g Cl_{2} with an excess of P_{4}.
Having identified Cl_{2} as the limiting reactant, we can complete the calculation as follows, using a stepwise approach.
Convert from grams of Cl_{2} to moles of Cl_{2} by using the molar mass of Cl_{2}. ? mol Cl_{2} = 323 g Cl_{2} × \frac{1 mol Cl_{2}}{70.91 g Cl_{2}} = 4.56 mol Cl_{2}
Convert from moles of Cl_{2} to moles of PCl_{3} by using the stoichiometric factor. ? mol PCl_{3} = 4.56 mol Cl_{2} × \frac{4 mol PCl_{3}}{6 mol Cl_{2}} = 3.04 mol PCl_{3}
Convert from moles of PCl_{3} to grams of PCl_{3} by using the molar mass of PCl_{3}. ? g PCl_{3} = 3.04 mol PCl_{3} × \frac{137.3 g PCl_{3}}{1 mol PCl_{3}} = 417 g PCl_{3}
Using the conversion pathway approach, we combine the steps into a single line calculation.
? g PCl_{3} = 323 g Cl_{2} × \frac{1 mol Cl_{2}}{70.91 g Cl_{2}} × \frac{4 mol PCl_{3}}{6 mol Cl_{2}} × \frac{137.3 g PCl_{3}}{1 mol PCl_{3}} = 417 g PCl_{3}
Assess
A different approach that leads to the same result is to do two separate calculations. First, calculate the mass of PCl_{3} produced by the reaction of 323 g Cl_{2} with an excess of P_{4} (= 417 g PCl_{3}), and then calculate the mass of PCl_{3} produced by the reaction of 125 g P_{4} with an excess of Cl_{2} (= 554 g PCl_{3}). Only one answer can be correct, and it must be the smaller of the two. An advantage of this approach is that it can be easily generalized to deal with reactions involving many reactants.
