Question

$27gmAl$ is heated with $49ml$ of $H_{2}S{O}_4$ (specific gravity $=2$) produces $H_2$ gas. Percentage of $Al$ reacted with $H_{2}S{O}_4$ is:

• A. $50\%$
• B. $66.66\%$
• C. $75\%$
• D. $80\%$

Answer 1

The correct option is B $66.66\%$

$2Al+3H_{2}S{O}_4⟶A{l}_2{\left(S{O}_4\right)}_3+3H_2\uparrow$

$2$ moles of $Al$ reacts with $3$ moles of $H_{2}S{O}_4$.

i.e. $2\times 27g$ reacts with $3\times 98g$ $H_{2}S{O}_4$.

Mass of $H_{2}S{O}_4$ present $=49\times 2=98g$

$\therefore$ Mass of $Al$ reacted $=\frac{2}{3}\times 27=18g$

$\therefore \%$ $Al$ reacted $=\frac{18}{27}\times 100=66.66\%$