If you start with 12 g of P2O5 and 71 g of H2O, which would be the limiting reactant?

Question

2P2O5+6H2O—>4H3PO4

if i got an answer to this problem it would really help

Answer 1

The chemical reaction is

2 P₂O₅ + 6 H₂O —> 4 H₃PO₄

You cannot tell from the mass what the limiting reactant will be. The limiting reactant is the reactant that produces the LEAST amount of product.

Therefore, you must convert the mass to moles, and calculate how much product (H₃PO₄) will be produced with each reactant.

12 g P₂O₅ × (1 mol P₂O₅ / 283.9 g P₂O₅) = 0.04227 mol P₂O₅

0.04227 mol P₂O₅ × (4 mol H₃PO₄ / 2 mol P₂O₅) = 0.08454 mol H₃PO₄

71 g H₂O × (1 mol H₂O / 18.02 g H₂O) = 3.940 mol H₂O

3.940 mol H₂O × (4 mol H₃PO₄ / 6 mol H₂O) = 2.627 mol H₃PO₄

Since P₂O₅ produces the least amount of H₃PO₄, P₂O₅ is the limiting reactant in this problem.

Answer 2