\(\text{a) }n_{H_2O}=\dfrac{m}{M}=\dfrac{0,9}{18}=0,05\left(mol\right)\)
\(pthh:CuO+H_2\overset{t^o}{\rightarrow}Cu+H_2O\left(1\right)\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }x\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }x\\ PbO+H_2\overset{t^o}{\rightarrow}Pb+H_2O\left(2\right)\\ \text{ }\text{ }\text{ }y\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }y\)
Từ (1) và (2), ta có hệ phương trình:
\(\left\{{}\begin{matrix}x+y=0,05\\80x+217y=8,59\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{113}{6850}\\y=\dfrac{459}{13700}\end{matrix}\right.\)
\(\Rightarrow m_{CuO}=n\cdot M=80\cdot x=80\cdot\dfrac{113}{6850}=1,32\left(g\right)\\ \Rightarrow m_{PbO}=n\cdot M=217\cdot\dfrac{459}{13700}=7,27\left(g\right)\)
\(\Rightarrow\%CuO=\dfrac{1,32\cdot100}{8,59}=15,37\%\\ \%PbO=\dfrac{7,27\cdot100}{8,59}=84,63\%\)
b) Theo \(pthh\left(1\right):n_{Cu}=n_{CuO}=\dfrac{113}{6850}\left(mol\right)\)
Theo \(pthh\left(2\right):n_{Pb}=n_{PbO}=\dfrac{459}{13700}\left(mol\right)\)
\(\Rightarrow m_{Cu}=n\cdot M=\dfrac{113}{6850}\cdot64=1,06\left(g\right)\\ m_{Pb}=n\cdot M=\dfrac{459}{13700}\cdot207=6,94\left(g\right)\\ \Rightarrow m_{h^2}=1,06+6,94=8\left(g\right)\)
\(\Rightarrow\%Cu=\dfrac{1,06\cdot100}{8}=13,25\%\\ \%Pb=\dfrac{6,94\cdot100}{8}=86,75\%\\ \)
