You have 8.32 g of P4(s) and 3.22 g of Cl2. Considering the following unbalanced equation, how much PCl3 will be produced: P4 + Cl2 -> PCl3

Question

You have 8.32 g of P4(s) and 3.22 g of Cl2. Considering the following unbalanced equation, how much PCl3 will be produced: P4 + Cl2 -> PCl3

(A) 4.12 g

(B) 1.82 g

(C) 0.27 g

(D) 15.78 g

Answer 1

First, we must write a correctly balanced equation:

P₄ + 6Cl₂ ==> 4PCl₃ .. balanced equation

Next, since we are given amounts of BOTH reactants, we must find which, if either, is limiting. One way to do this is to divide mols of each reactant by the corresponding coefficient in the balanced equation.

P₄: 8.32 g x 1 mol P₄ / 123.9 g = 0.0671 mols (÷1->0.067)

Cl₂: 3.22 g x 1 mol Cl₂ / 70.91 = 0.0454 mols (÷6->0.0076)

Since 0.0076 is < 0.067, this indicates that Cl2 is limiting, and mols of Cl2 (0.0454) will determine amount of product formed.

Mass PCl₃ = 0.0454 mols Cl₂ x 4 mols PCl₃ / 6 mols Cl₂ x 137 g PCl₃/mol = 4.14 g PCl₃ (Answer A)