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Q.
What weights of $P_{4}O_{6}$ and $P_{4}O_{10}$ will be produced by the combustion of $31\,g$ of $P_{4}$ in $32\,g$ of oxygen leaving no $P_{4}$ and $O_{2}$ .
NTA AbhyasNTA Abhyas 2022
Solution:
$P _4+ O _2 \rightarrow P _4 O _6+ P _4 O _{10} \text { (Non-balanced) }$
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 $=31$ $\left.g \,mol ^{-1}\right)$
$31=a \times 4 \times 31+b \times 4 \times 31 $
$4 a+4 b=1 \ldots \ldots \ldots(1)$
Similarly,
Initial weight of oxygen = Final weight of oxygen
$32= a \times 6 \times 16+ a \times 10 \times 16$
$3 a+5 b=1 \ldots \ldots \ldots(2)$
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 $284\, g$ $mol ^{-1}$, respectively.
So weight of $P _4 O _6=\frac{1}{8} \times 220=27.5\, g $
$ P _4 O _{10}=\frac{284}{8}=35.5 \,g .$