Explanation:
\(3CO(g)+Fe_2O_3(s)\rightarrow 2Fe(s)+3CO_2(g)\)
1)Mass of CO when 210.3 g of Fe produced.
Number of moles of \(Fe\) in 210.3 g=
\(\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}\)
\(=\frac{210.3}{55.84 g/mol}=3.76 mol\)
According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.76 moles of Fe will be obtained from : \(\frac{3}{2}\times 3.76 moles\) of CO that is 5.64 moles.
Mass of CO in 5.64 moles =
\(\text{Number of moles}\times \text{molar mass of CO}=5.46\times 28 g/mol=157.92 g\)
2)Mass of CO when 209.7 g of Fe produced.
Number of moles of \(Fe\) in 209.7 g=
\(\frac{\text{Given mass of Fe}}{\text{Molar mass of Fe}}\)
\(=\frac{209.7}{55.84 g/mol}=3.75 mol\)
According to reaction, 2 moles of Fe are obtained from 3 moles of CO, then 3.75 moles of Fe will be obtained from : \(\frac{3}{2}\times 3.75 moles\) of CO that is 5.625 moles.
Mass of CO in 5.625 moles =
\(\text{Number of moles}\times \text{molar mass of CO}=5.625\times 28 g/mol=157.5 g\)
