Bài 2:
CuO + H2 → Cu + H2O
\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
\(n_{Cu}=\dfrac{5,6}{64}=0,0875\left(mol\right)\)
Theo bài: \(n_{CuO}=n_{Cu}\)
Theo bài: \(n_{CuO}>n_{Cu}\)
\(\Rightarrow CuOdư\)
Theo PT: \(n_{CuO}pư=n_{Cu}=0,0875\left(mol\right)\)
\(\Rightarrow m_{CuO}pư=0,0875\times80=7\left(g\right)\)
\(\Rightarrow H=\dfrac{7}{8}\times100\%=87,5\%\)