# How many grams of diphosphorus trioxide, P_2O3, are required to produce 10.2 moles of phosphorous acid, H_3PO_3 in the reaction P_2O_3 + 3H_2O -> 2H_3PO_3?

Feb 15, 2017

$\text{561 g P"_2"O"_3}$ are required to produce $\text{10.2 mol H"_3"PO"_3}$.

#### Explanation:

Balanced Equation
$\text{P"_2"O"_3 + "3H"_2"O}$$\rightarrow$$\text{2H"_3"PO"_3}$

First use the balanced equation to determine the mole ratio between $\text{H"_3"PO"_3}$ and $\text{P"_2"O"_3}$. This ratio will be used to determine the moles $\text{P"_2"O"_3}$ required to produce $10.2$ moles $\text{H"_3"PO"_3}$.

Mole Ratio Between $\text{P"_2"O"_3}$ and $\text{H"_3"PO"_3}$ from the balanced equation.

$\left(1 {\text{mol P"_2"O"_3)/(2"mol H"_3"PO}}_{3}\right)$ and $\left(2 {\text{mol H"_3"PO"_3)/(1"mol P"_2"O}}_{3}\right)$

Multiply the moles $\text{H"_3"PO"_3}$ by the molar mass that cancels $\text{H"_3"PO"_3}$ and leaves $\text{P"_2"O"_3}$

10.2color(red)cancel(color(black)("mol H"_3"PO"_3))xx(1"mol P"_2"O"_3)/(2color(red)cancel(color(black)("mol H"_3"PO"_3)))="5.10 mol P"_2"O"_3

Now that the moles $\text{P"_2"O"_3}$ required to produce $\text{10.2 mol H"_3"PO"_3}$ are known, multiply the number of moles by its molar mass, $\text{109.945 g/mol}$. https://www.ncbi.nlm.nih.gov/pccompound?term=P2O3 This will give the mass in grams needed for $\text{P"_2"O"_3}$ to produce $\text{10.2 mol H"_3"PO"_3}$.

5.10color(red)cancel(color(black)("mol P"_2"O"_3))xx(109.945"g P"_2"O"_3)/(1color(red)cancel(color(black)("mol P"_2"O"_3)))="561 g P"_2"O"_3 rounded to three significant figures