Mg + 2HCl → MgCl2 + H2
Cu + HCl → X
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a) Theo PT: \(n_{HCl}=2n_{H_2}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)
b) Theo PT: \(n_{MgCl_2}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\)