Câu 17:

PTHH: $2\text{Al} + \text{Fe}_2\text{O}_3 \overset{\text{t}^\circ}{\rightarrow} \text{Al}_2\text{O}_3 + 2\text{Fe}$

$\text{n}_{\text{Al ban đầu}} = \dfrac{10{,}8}{27} = 0{,}4\text{ mol}$

$\text{n}_{\text{Fe}_2\text{O}_3\text{ ban đầu}} = \dfrac{48}{160} = 0{,}3\text{ mol}$

Ta có: $\dfrac{0{,}4}{2} < \dfrac{0{,}3}{1}$ nên hiệu suất phản ứng tính theo $\text{Al}$

$\text{m}_{\text{Al phản ứng}} = 10{,}8 - 2{,}7 = 8{,}1\text{ g}$

$\text{H} = \dfrac{\text{m}_{\text{Al phản ứng}}}{\text{m}_{\text{Al ban đầu}}} \cdot 100\% = \dfrac{8{,}1}{10{,}8} \cdot 100\% = 75\%$