Question

Calculate the amount of $H_{2}S{O}_4$ produced (in g) when $40$ mL $H_{2}O$ (d=$0.9$ g/mL) reacts with $4926$ L $S{O}_3$ at $1$ atm. and $300$ K according to the following reaction, $H_{2}O+S{O}_3\to H_{2}S{O}_4$.

• A. $196$ g
• B. $200$ g
• C. $214$ g
• D. $225$ g

Answer 1

The correct option is A $196$ g

$H_{2}O+S{O}_3\to H_{2}S{O}_4$
Mass of $H_{2}O$ given $=40\times 0.9=36$ g
Moles of $H_{2}O$ given $=\frac{36}{2}=18$ g
Moles of $S{O}_3$ given $=\frac{4926}{22.4}=219.9$ mol
Here, water is limiting reagent and moles of $H_{2}S{O}_4$ produced is $2$ mol.
Mass of $H_{2}S{O}_4=2\times 98=196$ g.