\(n_{CuSO_4}=0,2.0,5=0,1\left(mol\right);n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\) (1)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (2)
\(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\) (3)
\(Fe+2HCl\rightarrow FeCl_2+H_2\) (4)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) (5)
\(Fe+CuSO_4\rightarrow FeSO_4+Cu\) (6)
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\Rightarrow27x+56y=11\left(\text{*}\right)\)
Theo PT (1), (2), (3), (4), (5), (6): \(n_{CuSO_4}+n_{H_2}=n_{Fe}+\dfrac{3}{2}n_{Al}\)
=> \(\dfrac{3}{2}x+y=0,1+0,3=0,4\left(\text{*}\text{*}\right)\)
Từ (*), (**) => a = 0,2; b = 0,1
=> \(\%m_{Al}=\dfrac{0,2.27}{0,2.27+0,1.56}.100\%=49,1\%\)