\(n_{H2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
a) \(n_{Mg}=n_{H2}=0,2\left(mol\right)\rightarrow m_{Mg}=0,2.24=4,8\left(g\right)\)
b) \(n_{H2}=2n_{HCl}=0,4\left(mol\right)\rightarrow V_{ddHCl}=\dfrac{0,4}{0,5}=0,8\left(l\right)\)