mCuO = \(50\times\dfrac{20}{100}=10\left(g\right)\) => nCuO = \(\dfrac{10}{80}=0,125\) mol
mFeO = 50 - 10 = 40 (g) => nFeO = \(\dfrac{40}{72}=0,6\) mol
Pt: CuO + H2 --to--> Cu + H2O
0,125 ---> 0,125..........................(mol)
.....FeO + ...H2 --to--> Fe + H2O
0,6 mol-> 0,6 mol
VH2 = (0,125 + 0,6) . 22,4 = 16,24 (lít)
mCuO=20%.50=10(g)
=>nCuO=10/80=0,125(mol)
mFeO=50-10=40(g)
=>nFeO=40/72=0,6(mol)
CuO+H2----t*--->Cu+H2O
0,125__0,125
FeO+H2--->Fe+H2O
0,6___0,6
\(\Sigma nH_2\)=0,125+0,6=0,725(mol)
=>VH2=0,725.22,4=16,24(l)