a) Gọi $n_{CuO} = a(mol) ; n_{MgO} = b(mol)$
$\Rightarrow 80a + 40b = 10(1)$
$CuO + 2HCl \to CuCl_2 + H_2O$
$MgO + 2HCl \to MgCl_2 + H_2O$
$n_{HCl} = 2a + 2b = 0,1.3 = 0,3(2)$
Từ (1)(2) suy ra a = 0,1 ; b = 0,05
$\%m_{CuO} = \dfrac{0,1.80}{10}.100\% = 80\%$
$\%m_{MgO} = 100\% - 80\% = 20\%$
b)
$C_{M_{CuCl_2}} = \dfrac{0,1}{0,1} = 1M$
$C_{M_{MgCl_2}} = \dfrac{0,05}{0,1} = 0,5M$
\(n_{HCl}=0.01\cdot3=0.03\left(mol\right)\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
a 2a a a
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
b 2b b b
Theo đề,ta có: \(\left\{{}\begin{matrix}80a+40b=10\\2a+2b=0.3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0.1\\b=0.2\end{matrix}\right.\)
\(\%m_{CuO}=\dfrac{0.1\cdot80}{10}=\dfrac{8}{10}=80\%\)
\(\%m_{MgO}=20\%\)
