$^\text{n}$ $\text{H$_2$}$ = $\dfrac{2,24}{22,4}$ = $0,1$ $\text{mol}$
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$\text{Fe + 2HCl $\longrightarrow$ FeCl$_2$ + H$_2$}$
$\text{0,1 mol 0,1 mol}$
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$\text{%$^\text{m}$ Fe = $\dfrac{^\text{m}\text{Fe}}{^\text{m}\text{ hỗn hợp}}$ . 100% = $\dfrac{0,1 . 56}{20}$ . 100% = 28%}$
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$\text{⇒ %$^\text{m}$ ZnO = 100% - 28% = 72%}$
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$\text{⇒ Chọn B}$