PTHH: \(CuO+H_2\rightarrow Cu+H_2O\)
\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
a. Theo PT ta có: \(n_{CuO}=n_{Cu}=0,2\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,2.64=12,8\left(g\right)\)
b. Theo PT ta có: \(n_{CuO}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)