Question

A flask is of one litre capacity containing $N{H}_3$ at 1 $atm$ and ${25}^{o}C$. A spark is passed through until all the $N{H}_3$ is decomposed into $N_2$ and $H_2$.The moles of $H_2$ formed are $6.1305\times {10}^{-x}$.
The value of $x$ is...

Answer 1

$N_2+3H_2\to 2N{H}_3$
Mole of $N_2$ formed $=\left(\frac{1}{2}\right)\times moleofN{H}_3$
$=\left(\frac{1}{2}\right)\times 4.087\times {10}^{-2}$
$=2.0435\times {10}^{-2}$
and mole of $H_2$ formed $=\left(3/2\right)\times$ $moleof$ $N{H}_3$
$=\left(3/2\right)\times 4.087\times {10}^{-2}$
$=6.1305\times {10}^{-2}$
so value of $x$ is 2.