How many grams of H3PO4 are produced when 12.81 moles of water react with excess P4O10? Select one: a. 837 grams b. 1880 grams c. 471 grams d. 8.54 grams

To determine how many grams of H₃PO₄ (phosphoric acid) are produced from the reaction of water (H₂O) with an excess of P₄O₁₀ (phosphorus pentoxide), we'll first need to write down the balanced chemical equation for the reaction:

P₄O₁₀ + 6H₂O → 4H₃PO₄

From this equation, we can see that 1 mole of P₄O₁₀ reacts with 6 moles of H₂O to produce 4 moles of H₃PO₄.

You have started with 12.81 moles of water. According to the reaction stoichiometry, 6 moles of water produce 4 moles of H₃PO₄. To find out how many moles of H₃PO₄ are produced from 12.81 moles of H₂O, you would set up a proportion:

\[ 6 \text{ moles H₂O} : 4 \text{ moles H₃PO₄} :: 12.81 \text{ moles H₂O} : x \text{ moles H₃PO₄} \]

Let's solve for x (moles of H₃PO₄):

\[ x = \frac{4 \text{ moles H₃PO₄} \times 12.81 \text{ moles H₂O}}{6 \text{ moles H₂O}} \] \[ x = \frac{4 \times 12.81}{6} \] \[ x = \frac{51.24}{6} \] \[ x = 8.54 \text{ moles H₃PO₄} \]

Now, we'll need to convert moles of H₃PO₄ to grams. The molar mass of H₃PO₄ is approximately 98 g/mol. We'll multiply the moles of H₃PO₄ by the molar mass to find the grams:

\[ 8.54 \text{ moles H₃PO₄} \times 98 \text{ g/mol} = 837.32 \text{ grams H₃PO₄} \]

So, when 12.81 moles of water react with an excess of P₄O₁₀, approximately 837 grams of H₃PO₄ are produced.

Therefore, the correct answer is a. 837.