Gọi số mol Fe, Zn là a, b (mol)
=> 56a + 65b = 2,33 (1)
\(n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)
PTHH: Fe + H2SO4 --> FeSO4 + H2
a------------------------>a
Zn + H2SO4 --> ZnSO4 + H2
b------------------------>b
=> a + b = 0,04 (2)
(1)(2) => a = 0,03 (mol); b = 0,01 (mol)
=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,03.56}{2,33}.100\%=72,1\%\\\%m_{Zn}=\dfrac{0,01.65}{2,33}.100\%=27,9\%\end{matrix}\right.\)