How many grams of H3PO4 are produced when 10.0 moles of water react with an excess of P4O10?

P4O10 (s) + 6H2O (l) → 4H3PO4 (aq)

652.6 g of H₃PO₄ are formed

Explanation:

Reaction is:

P₄O₁₀ (s) + 6H₂O (l) → 4H₃PO₄ (aq)

As P₄O₁₀ is the reactant in excess, the limiting is water.

Ratio is 6:4. We make a rule of three:

6 moles of water produce 4 moles of phosphoric acid

Therefore, 10 moles of water may prodyce (10.4) / 6 = 6.66 moles of H₃PO₄

We convert the moles to mass:

6.66 moles. 97.9 g / 1mol = 652.6 g