Hòa tan hoàn toàn 13,8 gam hỗn hợp X gồm Al, Fe vào dung dịch H2SO4 loãng, thu được 11,1555 lít khí H2 (đkc).

Đáp án đúng là: C

$n_{H_2}\text{ = 0,45 mol}$

$\left.\begin{array}{l}\text{Fe (a) }\stackrel{}{\to }{\text{ Fe}}^{+2}\text{ + 2e (2a) }\\ Al\text{ (b) }\stackrel{}{\to }{\text{ Al}}^{+3}\text{ + 3e (3b)}\end{array}\right|\left|\begin{array}{l}2H^{+}\text{ + 2.1e }\stackrel{}{\to }{\text{ H}}_2\\ \text{ 0,9 }\leftarrow \text{ 0,45}\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}\stackrel{BT\text{ e}}{\to }\text{ 2a + 3b = 0,9}\\ \stackrel{m_{kl}}{\to }\text{ 56a + 27b = 13,8}\end{array}\right.\Rightarrow \left\{\begin{array}{l}\text{a = 0,15}\\ \text{b = 0,2}\end{array}\right.\stackrel{}{\to }{\text{ \%m}}_{Al}\text{ = 39,13\%}$