$m_{Fe_2O_3} = 60.80\% = 48(gam) \Rightarrow n_{Fe_2O_3} = \dfrac{48}{160} = 0,3(mol)$
$m_{CuO} = 60 - 48 = 12(gam) \Rightarrow n_{CuO} = \dfrac{12}{80} = 0,15(mol)$
$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
$n_{H_2} = 3n_{Fe_2O_3} + n_{CuO} = 1,05(mol)$
$V_{H_2} = 1,05.22,4 = 23,52(lít)$
\(m_{Fe_2O_3}=80\%.60=48\left(g\right)\Rightarrow n_{Fe_2O_3}=0,3\left(mol\right)\)
\(m_{CuO}=60-48=12\left(g\right)\Rightarrow n_{CuO}=0,15\left(mol\right)\)
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
0,3..............0,9
\(CuO+H_2\rightarrow Cu+H_2O\)
0,15.........0,15
=> \(V_{H_2}=\left(0,9+0,15\right).22,4=23,52\left(g\right)\)
