Question

What weights of $P_4{O}_6$ and $P_4{O}_{10}$ will be produced by the combustion of $31g$ of $P_4$ in $32g$ of oxygen leaving no $P_4$ and $O_2$ .

• A. $2.75g,219.5g$
• B. $27.5g,35.5g$
• C. $55g,71g$
• D. $17.5g,190.5g$

Answer 1

Answer: (B)

$P_4+O_2\to P_4{O}_6+P_4{O}_{10}\text{ (Non-balanced) }$
Given that $31$ gram of $P_4$ and $32$ gram of $O_2$ react completely.
According to question, weight of $P$ is conserved.
Let mole of $P_4{O}_6=a$ and
Mole of $P_4{O}_{10}=b$
Initial weight of $P=$ Final weight of $P$
( atomic mass $=31$ $gmo{l}^{-1})$
$31=a\times 4\times 31+b\times 4\times 31$
$4a+4b=1\dots \dots \dots (1)$
Similarly,
Initial weight of oxygen = Final weight of oxygen
$32=a\times 6\times 16+a\times 10\times 16$
$3a+5b=1\dots \dots \dots (2)$
From equation (1) and (2)
$b=\frac{1}{8}$ and $a=\frac{1}{8}$
Molecular mass of $P_4{O}_6$ and $P_4{O}_{10}$ are $220$ and $284g$ $mo{l}^{-1}$, respectively.
So weight of $P_4{O}_6=\frac{1}{8}\times 220=27.5g$
$P_4{O}_{10}=\frac{284}{8}=35.5g.$