\(n_{MgO}=\dfrac{1,6}{40}=0,04\left(mol\right)\)
\(n_{HCl}=0,2.4=0,8\left(mol\right)\)
PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
Xét tỉ lệ: \(\dfrac{0,04}{1}< \dfrac{0,8}{2}\) => MgO hết, HCl dư
PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,04--->0,08-----.0,04
\(\left\{{}\begin{matrix}C_{M\left(MgCl_2\right)}=\dfrac{0,04}{0,2}=0,2M\\C_{M\left(HCl.dư\right)}=\dfrac{0,8-0,08}{0,2}=3,6M\end{matrix}\right.\)
\(n_{MgO}=\dfrac{1,6}{40}=0,04\left(mol\right)\\
n_{HCl}=0,2.4=0,8\left(mol\right)\\
pthh:MgO+2HCl\rightarrow MgCl_2+H_2O\)
\(LTL:\dfrac{0,04}{1}< \dfrac{0,8}{2}\)
-> HCl dư
\(n_{HCl\left(p\text{ư}\right)}=2n_{MgO}=0,08\left(mol\right)\\
n_{MgCl_2}=n_{MgO}=0,04\left(mol\right)\\
C_{M\left(HCl\left(d\right)\right)}=\dfrac{0,8-0,04}{0,2}=3,8M\\
C_{M\left(MgCl_2\right)}=\dfrac{0,04}{0,2}=0,2M\)