\(n_{Mg}=\dfrac{4,8}{24}=0,2mol\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2
\(m_{HCl\left(líthuyết\right)}=0,4\cdot36,5=14,6g\)
\(\Rightarrow m_{HCl\left(cầndùng\right)}=\dfrac{14,6}{10,95}\cdot100=\dfrac{400}{3}g\approx133,3g\)
\(V_{H_2}=0,2\cdot22,4=4,48l\)