PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
a) Ta có: \(n_{HCl}=0,2\cdot1=0,2\left(mol\right)\)
\(\Rightarrow n_{MgO}=0,1mol\) \(\Rightarrow m_{MgO}=0,1\cdot40=4\left(g\right)\)
b) Theo PTHH: \(n_{MgO}=n_{MgCl_2}=0,1mol\)
\(\Rightarrow m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\)
c) Ta có: \(C_{M_{MgCL_2}}=\frac{0,1}{0,2}=0,5\left(M\right)\) (Coi như thể tích dd không đổi)
d) PTHH: \(2HCl+Ca\left(OH\right)_2\rightarrow CaCl_2+H_2O\)
Theo PTHH: \(n_{Ca\left(OH\right)_2}=\frac{1}{2}n_{HCl}=0,1mol\)
\(\Rightarrow m_{Ca\left(OH\right)_2}=0,1\cdot74=7,4\left(g\right)\) \(\Rightarrow m_{ddCa\left(OH\right)_2}=\frac{7,4}{7,4\%}=100\left(g\right)\)
