\(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
PTHH: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
0,1<-------0,1
=> \(\%CuO=\dfrac{0,1.80}{18,2}.100\%=43,956\%\)
=> \(\%Al_2O_3=100\%-43,956\%=56,044\%\)
\(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
PTHH:\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(0,1< \)------------\(0,1\)
\(\Rightarrow\%CuO=\dfrac{0,1.80}{18,2}.100\%=43,956\%\)
\(\Rightarrow\%Al_2O_3=100\%-43,956\%=56,044\%\)