a) \(Mg+2HCl\rightarrow MgCl_2+H2\uparrow\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Đặt \(n_{Mg}=x\left(mol\right);n_{Al}=y\left(mol\right)\)
\(\rightarrow24x+24y=8,85\left(1\right)\)
Theo PT : \(n_{MgCl_2}=x\left(mol\right);n_{MgCl_3}=y\left(mol\right)\)
\(\rightarrow95x+135,5y=39,625\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow x=\dfrac{62}{355}\left(mol\right);y=\dfrac{49}{284}\left(mol\right)\)
\(\rightarrow m_{Mg}=\dfrac{62}{355}.24\approx4,19\left(g\right)\)
\(\rightarrow m_{Mal}=8,85-4,19=4,66\left(g\right)\)
b) Theo PT : \(\text{Σ}n_{H_2}=0,5\text{Σ}n_{HCl}=\dfrac{1231}{2840}\left(mol\right)\)
\(\rightarrow V_{H_2}=\dfrac{1231}{2840}.22,4\approx9,71\left(l\right)\)
