Question

Equal weights of phosphorus and oxygen are heated in a closed vessel producing $P_2{O}_3$ and $P_2{O}_5$ in a 1:1 mole ratio. Find the mole percentage of the excess component left.

Answer 1

Let, 1 mole of each of $P_2{O}_5$ and $P_2{O}_3$ is formed. So the mole ratio of the reacting P and O atoms is 4:8 or 1:2

Let the mass of oxygen and phosphorus atoms be $W$.

Therfore, moles of oxygen atoms = $\frac{W}{16}$
moles of phosphorus atoms = $\frac{W}{31}$

From the mole ratio,
2 moles of oxygen atoms combine with 1 mole of phosphorus atoms.
$\therefore$ $\frac{W}{16}$ moles of oxygen atoms combine with $\frac{W}{32}$ moles of phosphorus atoms.

So here phosphorus is the excess component.

So, moles of phosphorus remaining $=\frac{W}{31}-\frac{W}{32}$

Hence mole parcentage of the excess component $=\frac{(\frac{W}{31}-\frac{W}{32})}{\frac{W}{31}}\times 100=\frac{1}{32}\times 100=3.125\%$