Question

One litre of $O_2$ and one litre of $H_2$ are taken in a vessel of 2 litres capacity at NTP. The gases are made to combine to form water. If the mass of water formed is $8.03\times {10}^{-x}$ g, then the value of $x$ is:

Answer 1

$\underset{\underset{\text{Volume after reaction (litres)}}{\text{Volume before reaction (in litres)}}}{\text{The given reaction is:}}\underset{\underset{0}{1}}{2H_2+}\underset{\underset{0.5}{1}}{O_2}\underset{\underset{1}{0}}{\to 2H_{2}O}$

AT NTP, the mole of $H_{2}O$ formed $=$ mole of $H_2$ used$=\frac{PV}{RT}=\frac{1\times 1}{0.0821\times 273}=4.46\times {10}^{-2}$

$\therefore$ Mass of $H_{2}O$ formed $=4.46\times {10}^{-2}\times 18$$=8.03\times {10}^{-1}g$

So, value of $x$ is $1$.