\(n_{Mg}=\frac{7,2}{24}=0,3\left(mol\right);n_{O_2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: \(2Mg+O_2-^{t^o}\rightarrow2MgO\)
Theo đề: 0,3......0,1.........................(mol)
Lập tỉ lệ: \(\frac{0,3}{2}>\frac{0,1}{1}\)=> Sau phản ứng Mg dư, O2 phản ứng hết
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
\(Mg_{dư}+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(TheoPT:n_{H_2}=n_{Mg\left(dư\right)}=0,3-0,2=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(m_{ddHCl}=\frac{(0,2.2+0,1.2)36,5}{29,2\%}=75\left(g\right)\)
\(\Rightarrow m_{ddsaup.ứ}=0,1.24+0,2.40+75-0,1.2=85,2\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\frac{\left(0,2+0,1\right).95}{85,2}.100=33,45\%\)
