mFe2O3=20.60%=12(g)
=>nFe2O3=12/160=0,075(mol)
mCuO=20-12=8(g)
=>nCuO=8/80=0,1(mol)
pt:
Fe2O3 + 3H2 ---> 2Fe + 3H2O
0,075_____0,225___0,15
CuO + H2 ---> Cu + H2O
0,1____0,1____0,1
mFe=0,15.56=8,4(g)
mCu=0,1.64=6,4(g)
\(\Sigma nH2=\)0,225+0,1=0,325(mol)
=>VH2=0,325.22,4=7,28(l)