\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(n_{HCl}=\dfrac{100.10,95\%}{36,5}=0,3\left(mol\right)\)
PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\) => MgO hết, HCl dư
PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,1----->0,2------>0,1
\(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,1.95}{100+4}.100\%=9,13\%\\C\%_{HCl\left(dư\right)}=\dfrac{\left(0,3-0,2\right).36,5}{100+4}.100\%=3,51\%\end{matrix}\right.\)
\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(m_{HCl}=\dfrac{100.10,95\%}{100}=10,95\left(g\right)\)
=> \(n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,1 ----> 0,2 ----- > 0,1
LTL : \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\)=> MgO đủ , HCl dư
Chất sau pứ gồm HCl dư và MgCl2
\(m_{HCl\left(dư.sau.pứ\right)}=\left(0,3-0,2\right).36,5=3,65\left(g\right)\)
\(m_{MgCl_2}=0,1.95=9,5\left(g\right)\)
\(C\%_{ddHCl}=\dfrac{3,65.100}{4+100}=3,51\%\)
\(C\%_{MgCl_2}=\dfrac{9,5.100}{4+100}=9,13\%\)
