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