Question

Concentrated nitric acid oxidizes phosphorus to phosphoric acid according to the following equation:

$\mathrm{P}+5{\mathrm{HNO}}_3\left(\mathrm{conc}.\right)\to {\mathrm{H}}_3{\mathrm{PO}}_4+{\mathrm{H}}_2\mathrm{O}+5{\mathrm{NO}}_2$

If $9.3\mathrm{g}$ of phosphorus was used in the reaction, calculate:

(a) Number of moles of phosphorus taken? $\left[\mathrm{H}=1,\mathrm{N}=14,\mathrm{P}=31,\mathrm{O}=16\right]$

Answer 1

Step 1: Chemical reaction

• Phosphorus $\left(\mathrm{P}\right)$ undergoes a chemical reaction with concentrated nitric acid $\left({\mathrm{HNO}}_3\right)$ and results in the formation of phosphoric acid ${\mathrm{H}}_3{\mathrm{PO}}_4$, water $\left({\mathrm{H}}_2\mathrm{O}\right)$ and nitrogen dioxide $\left({\mathrm{NO}}_2\right)$.
• The given reaction is as follows:

$\mathrm{P}+5{\mathrm{HNO}}_3\left(\mathrm{conc}.\right)\to {\mathrm{H}}_3{\mathrm{PO}}_4+{\mathrm{H}}_2\mathrm{O}+5{\mathrm{NO}}_2$

Step 2: Given Information

• $9.3\mathrm{g}$ of phosphorus

Step 3: Calculation of moles of Phosphorus

Atomic weight of Phosphorus$\left(\mathrm{P}\right)=31\mathrm{g}$.

• Since, $31\mathrm{g}$ is the mass of $1\mathrm{mole}$ of phosphorous.
• Therefore, $9.3\mathrm{g}$ is the mass of $\frac{1\mathrm{mole}}{31\mathrm{g}}\times 9.3\mathrm{g}=0.3\mathrm{moles}$ of phosphorous.

Therefore, $0.3$ moles of phosphorus are taken.