CuO + H2 ->Cu + H2O
nCuO=0,12(mol)
Theo PTHH ta có:
nCuO=nH2=nCu=0,12(mol)
mCu=64.0,12=7,68(g)
VH2=22,4.0,12=2,688(lít)
a, PTHH: CuO+H2--->Cu+H2O
nCuO= \(\dfrac{9,6}{80}=0,12\) mol
Theo pt: nCu=nCuO= 0,12 mol
=> mCu= 0,12.64= 7,68 g
b, Theo pt: nH2=nCuO= 0,12 mol
=> VH2= 0,12.22,4= 2,688 l
