Đặt \(n_{Zn}=a\left(mol\right),n_{Fe}=b\left(mol\right)\rightarrow65a+56b=14,9\left(1\right)\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Ta có : \(n_{H2SO4}=\dfrac{24,5}{98}=0,25\left(mol\right)\rightarrow a+b=0,25\left(2\right)\)
(1),(2) \(\rightarrow a=0,1,b=0,15\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{14,9}.100\%=43,62\%\\\%m_{Fe}=100\%-43,62=56,38\%\end{matrix}\right.\)
b) \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,1------------------------------->0,1
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,15---------------------------->0,15
\(\rightarrow V_{H2\left(đkc\right)}=\left(0,1+0,15\right).24,79=6,1975\left(l\right)\)
