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