A mixture of 93.4 g of Cl 2 and 20.4 g of P reacts completely to form PCl 3 and PCl 5 . Find the mass of PCl 5 produced.

Question



Answer 1

Hello!

First, write the chemical equation and then balance it.

__Cl₂ + __P → __PCl₃ + __PCl₅

4Cl₂ + 2P → PCl₃ + PCl₅

Then, determine the limiting reactant.

93.4g Cl₂ (1 mol Cl₂ / 70.9 g Cl₂) (1 mol PCl₅ / 4 mol Cl₂) (208.22 g PCl₅ / 1 mol PCl₅) = 68.575 g PCl₅

20.4g P (1 mol P / 30.97 g P) (1 mol PCl₅ / 2 mol P) (208.22 g PCl₅ / 1 mol PCl₅) = 68.577 g PCl₅

Technically, the amount of Cl₂ used in the reaction leads to a lesser amount of PCl₅ compared to P if the reactants are completely consumed. However, due to sig figs, both reactants will lead to a final answer of 68.6 g of PCl₅ produced.