MgO+2HCl-----> MgCl2 +H2
a) n\(_{MgO}=\frac{8}{40}=0.2\left(mol\right)\)
Theo pthh
n\(_{HCl}=2n_{MgO}=0,4\left(mol\right)\)
C\(_{M\left(HCl\right)}=\frac{0,4}{0,2}=2\left(M\right)\)
b) Theo pthh
n\(_{MgCl2}=n_{MgO}=0,2\left(mol\right)\)
C\(_{M\left(MgCl2\right)}=\frac{0,2}{0,2}=1\left(M\right)\)