Question

Answer 1

a) Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)

PTHH: 
2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂ 

a------------------------------------->1,5a

Fe + H₂SO₄ ---> FeSO₄ + H₂ 

b------------------------------>b

b) Ta có hệ PT: \(\left\{{}\begin{matrix}27a+56b=0,83\\1,5a+b=\dfrac{0,56}{22,4}=0,025\end{matrix}\right.\Leftrightarrow a=b=0,01\)

=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,01.27}{0,83}.100\%=32,53\%\\\%m_{Fe}=100\%-32,53\%=67,47\%\end{matrix}\right.\)