\(n_{MgO}=\frac{1,5}{40}=0,0375\left(mol\right)\)
\(n_{HCl}=\frac{100.5,475}{100.36,5}=0,15\left(mol\right)\)
\(PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\)
(mol)____0,0375__0,075____0,0375_________
Tỉ lệ: \(\frac{0,15}{2}>\frac{0,0375}{1}\rightarrow\) HCl dư 0,15 - 0,075 = 0,075(mol)
\(C\%_{\text{dd}MgCl_2}=\frac{0,0375.95}{1,5+100}.100\%=3,51\left(\%\right)\)
\(C\%_{HCl.du}=\frac{36,5.0,075}{1,5+100}.100\%=2,7\left(\%\right)\)