a)
\(n_{Fe} = \dfrac{48}{56} = \dfrac{6}{7}(mol)\)
\(Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\)
Theo PTHH : \(n_{H_2} = \dfrac{3}{2}n_{Fe} = \dfrac{9}{7}(mol)\\ \Rightarrow V_{H_2} = \dfrac{9}{7}.22,4 = 28,8(lít)\)
b)
Ta có :
\(n_{Fe_2O_3} = 0,5n_{Fe} = \dfrac{3}{7}(mol)\\ \Rightarrow m_{Fe_2O_3} = \dfrac{3}{7}.160 = 68,57(gam)\)