\(n_{Mg}=0,2mol\), \(n_{HCl}=0,5mol\)
a,\(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\)
b.Có: \(\frac{0,2}{1}< \frac{0,5}{2}\) => Mg hết, HCl dư
\(n_{HCL\left(pư\right)}=2n_{Mg}=0,4mol\)
\(n_{HCldư}=0,5-0,4=0,1mol\)
=> \(m_{HCl\left(dư\right)}=0,1.36,5=3,65g\)
c,\(v_{H_2}=0,2.22,4=4,48l\)