a) Mg + 2HCl → MgCl2 + H2
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{H_2}=2\times0,4=0,8\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,8\times36,5=29,2\left(g\right)\)
b) Theo PT: \(n_{MgCl_2}=n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=0,4\times95=38\left(g\right)\)