a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, \(n_{Mg}=\dfrac{1,2}{24}=0,05\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{HCl}=2n_{Mg}=0,1\left(mol\right)\\n_{MgCl_2}=n_{H_2}=0,05\left(mol\right)\end{matrix}\right.\)
⇒ mHCl = 0,1.36,5 = 3,65 (g)
mMgCl2 = 0,05.95 = 4,75 (g)
c, VH2 = 0,05.22,4 = 1,12 (l)