Question

The volume of $HCl\left(g\right)$ at STP required to clean up all $N{H}_3$ molecules adsorbed on surface is:

• A. $0.175$ L
• B. $0.238$ L
• C. $0.432$ L
• D. $0.526$ L

Answer 1

The correct option is D $0.526$ L

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.
$n_{HCl}=0.0235$
$\therefore 1\times V=0.0235\times 0.0821\times 273$
$V_{HCl}\left(g\right)=0.526L$