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

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.