\(n_{MgO}=\dfrac{4}{40}=0,1mol\\ n_{HCl}=0,2.2=0,4mol\\ MgO+2HCl\rightarrow MgO+H_2O\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,4}{2}\Rightarrow HCl.dư\\ n_{MgCl_2}=n_{MgO}=0,1mol\\ n_{HCl,pư}=0,1.2=0,2mol\\ m_{MgCl_2}=0,1.95=9,5g\\ m_{HCl,dư}=\left(0,4-0,2\right).36,5=7,3g\)
Ta có: \(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(n_{HCl}=0,2.2=0,4\left(mol\right)\)
PT: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\), ta được HCl dư.
Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{H_2O}=n_{MgO}=0,1\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{MgO}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,1.95=9,5\left(g\right)\\m_{H_2O}=0,1.18=1,8\left(g\right)\\m_{HCl\left(dư\right)}=0,2.36,5=7,3\left(g\right)\end{matrix}\right.\)