\(Zn+2HCl-->ZnCl_2+H_2\)
0,2_____________________ 0,2
\(H_2+CuO--to->Cu+H_2O\)
0,15____0,15_________0,15
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
=> \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(n_{CuO}=0,15\left(mol\right)\)
=> CuO ht H2 dư rồi tự tính nhé