Question

Excess of $NaOH$ and $H_{2}S{O}_4$ react with $Al$ separately and liberate $H_2$ in the ratio of :

• A. $1:1$
• B. $2:1$
• C. $1:2$
• D. $2:3$

Answer 1

The correct option is A $1:1$

Two moles of $Al$ react with six moles of sodium hydroxide($NaOH$) to form two moles of sodium aluminate and three moles of hydrogen.

$2Al+6NaOH\to 2N{a}_{3}Al{O}_3+3H_2$

Similarly, two moles of $Al$ react with three moles of sulphuric acid($H_{2}S{O}_4$) to form one mole of aluminium sulphate and three moles of hydrogen.

$2Al+3H_{2}S{O}_4\to A{l}_2(S{O}_4{)}_3+3H_2$

Thus, the ratio of hydrogen obtained in the two reactions is $1:1$.