# How many grams of H_3PO_4 are produced when 10.0 moles of water react with an excess of P_4O_10?

Jun 23, 2017

$653$ ${\text{g H"_3"PO}}_{4}$

#### Explanation:

We'll first write the chemical equation for this reaction:

${\text{P"_4"O"_10(s) + 6"H"_2"O"(l) rarr 4"H"_3"PO}}_{4} \left(a q\right)$

Let's now find the relative number of moles of phosphoric acid (${\text{H"_3"PO}}_{4}$) using the coefficients of the equation, and the fact that $10.0$ $\text{mol H"_2"O}$ react:

10.0cancel("mol H"_2"O")((4color(white)(l)"mol H"_3"PO"_4)/(6cancel("mol H"_2"O"))) = color(red)(6.67 color(red)("mol H"_3"PO"_4

Now, using the molar mass of ${\text{H"_3"PO}}_{4}$ (calculated to be $98.00$ $\text{g/mol}$), let's find the theoretical mass of phosphoric acid produced:

color(red)(6.67)color(white)(l)cancel(color(red)("mol H"_3"PO"_4))((98.00color(white)(l)"g H"_3"PO"_4)/(1cancel("mol H"_3"PO"_4))) = color(blue)(653 color(blue)("mol H"_3"PO"_4

Thus, if the reaction goes to completion, there will be a yield of color(blue)(653 sfcolor(blue)("grams of phosphoric acid".