m Fe2O3=28.\(\dfrac{75}{100}\)=21g =>n Fe2O3=\(\dfrac{21}{160}\)=0,13125 mol
m CuO=28-21=7 g =>n CuO=\(\dfrac{7}{80}\)=0,0875 mol
Fe2O3+3H2->2Fe+3H2O
0,13125-0,39375-0,2625
CuO+H2-to>Cu+H2O
0,0875-0,0875-0,0875 mol
=>m Fe=0,2625.56=14,7g
=>m Cu=0,0875.64=5,6g
=>n H2=0,39375+0,0875=0,48125mol