a) PTHH: CuO + H2 \(\underrightarrow{t^o}\) Cu + H2O
b) nCuO = \(\frac{1,6}{80}=0,02\left(mol\right)\)
Theo PT: n\(H_2\) = nCuO = 0,02 (mol)
=> V\(H_2\) = 0,02.22,4 = 0,448 (l)
c) Theo PT: nCu = nCuO = 0,02 (mol)
=> mCu = 0,02.64 = 1,28 (g)
nCuO= 1.6/80=0.02 mol
CuO + H2 -to-> Cu + H2O
0.02___0.02____0.02
VH2= 0.02*22.4=0.448l
mCu= 0.02*64=1.28g