Question

What weight of ${\mathrm{HNO}}_3$ is required to convert $62\mathrm{gms}{\mathrm{P}}_4$ to ${\mathrm{H}}_3{\mathrm{PO}}_4$ in the given reaction?

$\mathrm{P}+{\mathrm{HNO}}_3\to {\mathrm{H}}_3{\mathrm{PO}}_4+{\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O}$

Answer 1

Step 1: Write the balanced reaction and determine the molar mass of ${\mathrm{P}}_4$ and ${\mathrm{HNO}}_3$.

$\mathrm{P}+20 {\mathrm{HNO}}_3\to {\mathrm{H}}_3{\mathrm{PO}}_4+20 {\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O}$

Molar mass of ${\mathrm{P}}_4=124\mathrm{g}/\mathrm{mol}$

Molar mass of ${\mathrm{HNO}}_3=63\mathrm{g}/\mathrm{mol}$

Step 2: Determine the mass of ${\mathrm{HNO}}_3$ reacted with $124\mathrm{g}/\mathrm{mol}{\mathrm{P}}_4$

$1\mathrm{mole}{\mathrm{P}}_4$ reacts with $20\mathrm{moles}$ of ${\mathrm{HNO}}_3$

$124\mathrm{g}/\mathrm{mol}{\mathrm{P}}_4$ reacts with $20\times 63\mathrm{g}$ of ${\mathrm{HNO}}_3$

Step 3: Calculate the mass of ${\mathrm{HNO}}_3$ reacted with $64\mathrm{g}/\mathrm{mol}{\mathrm{P}}_4$

$64\mathrm{g}/\mathrm{mol}{\mathrm{P}}_4$ reacts with $\frac{20\times 63\times 62}{124}=630\mathrm{g}$ of ${\mathrm{HNO}}_3$

Hence, the weight of ${\mathrm{HNO}}_3$ required to convert $62\mathrm{gms}{\mathrm{P}}_4$ to ${\mathrm{H}}_3{\mathrm{PO}}_4$ in the given reaction is $630\mathrm{g}$.