Question

In a particular experiment, $310g$ of phosphorous ($P_4$) reacted with excess of oxygen to form $P_4{O}_{10}$ in 75% yield. Then $P_4{O}_{10}$ formed reacted with $54$ g of water to give $H_{3}P{O}_4.$ Amount of $H_{3}P{O}_4$ obtained (in gm) is :
(Consider only two places after decimal for calculation)

Answer 1

$P_4+5O_2\to P_4{O}_{10}$
$P_4{O}_{10}+6H_{2}O\to 4H_{3}P{O}_4$

In the firest step, for 75% yield, moles of $P_4{O}_{10}$ obtained.
= moles of $P_4\times \frac{75}{100}$
$=\frac{310}{124}\times \frac{75}{100}$
$=1.875moles$

Again,
$P_4{O}_{10}+6H_{2}O\to 4H_{3}P{O}_4$
Moles of water given $=\frac{54}{18}=3$ moles
Actually amount of water required for 1.875 moles of $P_4{O}_{10}$ is
$=1.875\times 6$
$=11.25$ moles of water.
$\therefore$ Water is limiting agent.
$\therefore$ 6 moles of $H_{2}O$ produced 4 moles of $H_{3}P{O}_4$
3 moles of $H_{2}O$ will produce
$=\frac{4}{6}\times 3$ moles of $H_{3}P{O}_4$
$=2$ moles of $H_{3}P{O}_4$
Molar mass of $H_{3}P{O}_4=98g$
Therefore, amount of $H_{3}P{O}_4$ obtained = $2\times 98g=196g$