a. \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
PTHH : H2 + CuO -> Cu + H2O
0,1 0,1 0,1
Xét tỉ lệ \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\) => H2 đủ , CuO dư
b. \(m_{H_2O}=0,1.18=1,8\left(g\right)\)
c. \(m_{Cu}=0,1.64=6,4\left(g\right)\)
