\(n_{Mg}=\frac{4,8}{24}=0,2\left(mol\right)\)
PTHH: Mg + 2HCl ---------> MgCl2 + H2
Theo PT : n HCl =2nMg =0,4 (mol)
=> \(V_{HCl}=\frac{0,4}{2}=0,2\left(l\right)=200\left(ml\right)\)
b) mdd HCl = 1,12 . 200 = 224 (g)
\(\Rightarrow C\%_{HCl}=\frac{0,4.36,5}{224}.100=6,52\%\)