- Xét TN1:
\(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,04<--------------------0,06
- Xét TN2:
\(n_{SO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: \(Cu+2H_2SO_{4\left(đ\right)}\rightarrow CuSO_4+SO_2+2H_2O\)
0,4<----------------------------0,4
=> Trong m (g) hỗn hợp chứa \(\left\{{}\begin{matrix}Al:0,04\left(mol\right)\\Cu:0,2\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,04.27}{0,04.27+0,2.64}.100\%=7,78\%\\\%m_{Cu}=\dfrac{0,2.64}{0,04.27+0,2.64}.100\%=92,22\%\end{matrix}\right.\)