\(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right);n_{H_2O}=\dfrac{10,8}{18}=0,6\left(mol\right)\)
PTHH:
\(H_2O+C\xrightarrow[]{t^o}CO+H_2\\ 2H_2O+C\xrightarrow[]{t^o}CO_2+2H_2\)
\(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)
\(Fe_2O_3+3CO\xrightarrow[]{t^o}2Fe+3CO_2\)
Theo PT: \(3n_{Fe_2O_3}=n_{H_2O}+n_{CO}\)
=> nCO = 3.0,3 - 0,6 = 0,3 (mol)
Theo PT: \(n_{H_2}=n_{H_2O}=0,6\left(mol\right)\)
Theo PT: \(n_{H_2}=2n_{CO_2}+n_{CO}\)
=> \(n_{CO_2}=\dfrac{0,6-0,3}{2}=0,15\left(mol\right)\)
=> \(\%V_{CO_2}=\%n_{CO_2}=\dfrac{0,15}{0,15+0,3+0,6}.100\%=14,28\%\)