Ta có:
\(n_{CuO}=\frac{1,6}{80}=0,02\left(mol\right)\)
Gọi \(n_{CuO\left(pư\right)}:x\left(mol\right)\)
\(PTHH:CuO+H_2\rightarrow Cu+H_2O\)
___________x____________x____
\(m_{cr}=m_{CuO\left(dư\right)}+m_{Cu}=1,344\)
\(\Rightarrow80.\left(0,02-x\right)+64x=1,344\)
\(\Leftrightarrow x=0,016\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\frac{0,016.64}{1,344}.100\%=76,19\%\\\%m_{Cu\left(dư\right)}=100\%-76,19\%=23,81\%\end{matrix}\right.\)
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