Question

The mass of $N{H}_{4}Cl$ formed after desorption of $N{H}_3$ from activated charcoal surface on treatment with $HCl\left(g\right)$ is :

• A. $1.258$ g
• B. $0.0335$ g
• C. $0.0235$ g
• D. $0.0135$ g

Answer 1

The correct option is A $1.258$ g

Total moles of $N{H}_3$ adsorbed on surface $=\frac{1.415\times {10}^{22}}{6.023\times {10}^{23}}=0.0235$.

Moles of $N{H}_3$ before adsorption $=\frac{100\times 2}{1000}=0.2$.

So, moles of $N{H}_3$ left after adosrption $=$ moles of $HCl$ required $=0.2-0.0235=0.175$.
Also, moles of $HCl$ required to clean surface $=$ moles of $N{H}_3$ adsorbed.
$n_{HCl}=0.0235$
$\therefore 1\times V=0.0235\times 0.0821\times 273$
$V_{HCl}\left(g\right)=0.526L$

$N{H}_3+HCl\to N{H}_{4}Cl$
Also, moles of $N{H}_{4}Cl$ formed $=$ moles of $N{H}_3$ adsorbed $=0.0235$.
So, mass of $N{H}_{4}Cl$ formed = $0.0235\times 53.5=1.258$ g.