a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, \(n_{Mg}=\dfrac{1,2}{24}=0,05\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Mg}=0,1\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,1.36,5}{10\%}=36,5\left(g\right)\)
c, \(n_{H_2}=n_{MgCl_2}=n_{Mg}=0,05\left(mol\right)\)
Ta có: m dd sau pư = 1,2 + 36,5 - 0,05.2 = 37,6 (g)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,05.95}{37,6}.100\%\approx12,63\%\)