In this example problem,
we use dimensional analysis to find the limiting reagent and calculate the theoretical yield, percent yield, and excess reagent remaining.
Our problem reads, "125 grams P4 is combined with 384 grams Cl2 according to the reaction: P4 solid plus 6 Cl2 gas produces 4 PCl3 liquid.
Part A: calculate the theoretical yield of PCl3 in grams.
Part B: if 426 grams PCl3 is actually produced from the reaction, calculate the percent yield.
And part C: calculate the mass in grams of excess reactant remaining.
In this problem, we're given amounts of two different reactants so we need to determine which one runs out first; this is a limiting reagent problem.
To calculate the theoretical yield, we need to determine how much product can be produced from each amount of reactant.
Whichever produces less product is the limiting reagent, and the amount it can produce is the theoretical yield.
We'll use dimensional analysis to convert 125 grams P4 into Gramm's PCL three
We can use the molar mass of P4 to convert from grams of P4 to moles of P4.
Then we can use the coefficients in the balanced chemical equation to convert from moles of P4 to moles of phosphorus trichloride, PCl3.
Then we can use the molar mass of phosphorus trichloride to convert from moles PCl3 to grams PCl3.
Completing the calculation, we should get 554 grams PCl3.
Next we have to convert 384 grams chlorine gas to grams PCl3 so we know which reactant produces less product.
We can use the molar mass of chlorine gas to convert from grams Cl2 to moles Cl2.
Then we can use the coefficients in the balanced chemical equation to convert from moles Cl2 to moles PCl3.
Then we use the molar mass of PCl3 to convert from moles PCl3 to grams PCl3.
Completing the calculation, we should get 496 grams PCl3.
Since Cl2 produce less product, chlorine gas is the limiting reagent; the amount of product it produces, 496 grams PCl3, is the theoretical yield.
Next we want to calculate percent yield.
Our equation for percent yield is actual yield divided by theoretical yield times 100%.
We're told when the reaction is performed, 426 grams of PCl3 is actually produced; this is the actual yield.
We calculated the theoretical yields in part A, which is 496 grams PCl3.
Completing the calculation we get 85.9%.
To determine how much excess reactant remains,
we first need to determine how much of our excess reactant is needed to react with our limiting reagent.
This will tell us how much of the excess reactant is used in the reaction.
We start the dimensional analysis with our limiting reagent, 384 grams Cl2.
We need to convert this into grams of excess reagent, P4.
We use the molar mass of Cl2 to convert from grams of Cl2 to moles of Cl2.
We use the coefficients in the balanced chemical equation to convert from moles Cl2 to moles P4.
Then we convert from moles of P4 to grams P4 using the molar mass of P4.
Completing the calculation, we should get 112 grams P4.
This means 112 grams of P4 are required to completely react with our limiting reagent, Cl2.
To determine how much P4 is left over, we take 125 grams P4 subtract 112 grams of P4 to get 13.2 grams of P4 remaining.
We started with 125 grams of P4, we use 112 grams in the reaction, so there should be 13.2 grams of P4 remaining.