\(a.n_{HCl}=\dfrac{150.7,3\%}{36,5}=0,3\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{MgO}=a\left(mol\right)\\n_{CuO}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\ MgO+2HCl\rightarrow MgCl_2+H_2\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ \Rightarrow\left\{{}\begin{matrix}40a+80b=8\\2a+2b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \Rightarrow\%m_{CuO}=\dfrac{0,05.80}{8}.100=50\%\\ \Rightarrow\%m_{MgO}=100\%-50\%=50\%\\ b.m_{ddsau}=150+8=158\left(g\right)\\ C\%_{ddMgCl_2}=\dfrac{0,1.95}{158}.100\approx6,013\%\\ C\%_{ddCuCl_2}=\dfrac{135.0,05}{158}.100\approx4,272\%12\)
Pt: \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)
1 2 1 1
a 2a 0,1
\(CuO+2HCl\rightarrow CuCl_2+H_2O|\)
1 2 1 1
b 2b 0,05
a) Gọi a là số mol của MgO
b là số mol của CuO
\(m_{MgO}+m_{CuO}=8\left(g\right)\)
⇒ \(n_{MgO}.M_{MgO}+n_{CuO}.M_{CuO}=8g\)
⇒ 50a + 80b = 8g (1)
Ta có :
\(m_{ct}=\dfrac{7,3.500}{100}=10,95\left(g\right)\)
\(n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
⇒ 2a + 2b = 0,3(2)
Từ (1),(2), ta có hệ phương trình :
40a + 80b = 8g
2a + 3b = 0,3
⇒ \(\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)
\(m_{MgO}=0,1.40=4\left(g\right)\)
\(m_{CuO}=0,05.80=4\left(g\right)\)
0/0MgO = \(\dfrac{4.100}{8}=50\)0/0
0/0CuO = \(\dfrac{4.100}{8}=50\)0/0
b) Có : \(n_{MgO}=0,1\left(mol\right)\Rightarrow n_{MgCl2}=0,1\left(mol\right)\)
\(n_{CuO}=0,05\left(mol\right)\Rightarrow n_{CuCl2}=0,05\left(mol\right)\)
\(m_{MgCl2}=0,1.95=9,5\left(g\right)\)
\(m_{CuCl2}=0,05.135=6,75\left(g\right)\)
\(m_{ddspu}=8+150=158\left(g\right)\)
\(C_{MgCl2}=\dfrac{9,5.100}{158}=6,1\)0/0
\(C_{CuCl2}=\dfrac{6,75.100}{158}=4,27\)0/0
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