\(a,n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ n_{HCl}=1.V=V\left(mol\right);n_{H_2SO_4}=0,75.V\left(mol\right)\\ \Rightarrow n_{H\left(trong.axit\right)}=V+0,75.V.2=2,5V\left(mol\right)=2.n_{H_2}=2.n_{Mg}=2.0,15=0,3\left(mol\right)\\ \Leftrightarrow V=\dfrac{0,3}{2,5}=0,12\left(lít\right)\\ b,n_{H_2}=n_{Mg}=0,15\left(mol\right)\Rightarrow V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)
\(\dfrac{C_{M_{HCl}}}{C_{M_{H_2SO_4}}}=\dfrac{1}{0.75}=\dfrac{4}{3}\)
\(n_{HCl}=a\left(mol\right)\Rightarrow n_{H_2SO_4}=\dfrac{3a}{4}\left(mol\right)\)
\(n_{Mg}=\dfrac{3.6}{24}=0.15\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(0.5a......a\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(\dfrac{3a}{4}.....\dfrac{3a}{4}\)
\(n_{Mg}=0.5a+\dfrac{3a}{4}=1.25a=0.15\left(mol\right)\)
\(\Rightarrow a=0.12\)
\(V=\dfrac{0.12}{1}=0.12\left(l\right)=120\left(ml\right)\)
\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)