\(n_{Mg}=\frac{2,4}{24}=0,1\left(mol\right)\)
\(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\)
(mol)______0,1____0,2________0,1__0,1__
\(V_{H_2}=22,4.0,1=2,24\left(l\right)\)
\(m_{ddHCl}=\frac{m_{ct}.100}{C\%}=\frac{\left(36,5.0,2\right).100}{100}=7,3\left(g\right)\)