Trong 10 tấn thép: \(m_{Fe}=9,8\)(tấn)
\(Fe_2O_3\left(0,0875\right)+3H_2\left(0,2625\right)-t^o->2Fe\left(0,175\right)+3H_2O\)
\(n_{Fe}=0,175\)(tấn mol)
Theo PTHH: \(n_{Fe_2O_3}=0,0875\)(tấn mol)
\(m_{Fe_2O_3}=0,0875.160=14\)(tấn)
\(n_{H_2}=0,2625\)(tấn mol)\(=262500\left(mol\right)\)
\(\Rightarrow V_{H_2}\left(đktc\right)=5880000\left(l\right)\)