a: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,2 0,4 0,2 0,2
\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)
=>\(n_{H_2}=0.2\left(mol\right)\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(lít\right)\)
b:
\(n_{MgCl_2}=n_{Mg}=0.2\left(mol\right)\)
\(m_{MgCl_2}=0.2\left(24+35.5\cdot2\right)=19\left(g\right)\)
c: \(C\%\left(HCl\right)=\dfrac{0.4\cdot36.5}{100}=14.6\%\)