Answer:
[tex]m_{CO}^{leftover}=3.864gCO[/tex]
Explanation:
Hello,
In this case, the first step is to compute the moles of the reacting both iron (III) oxide and carbon monoxide as follows:
[tex]n_{Fe_2O_3}=22.95gFe_2O_3*\frac{1molFe_2O_3}{160gFe_2O_3}=0.1434molFe_2O_3\\ n_{CO}=15.91molCO*\frac{1molCO}{28gCO}=0.5682molCO[/tex]
Now, we can take the reacting iron (III) oxide's moles in order to compute the actual consumed carbon monoxide's moles that would completely react with the iron (III) oxide's moles as follows:
[tex]n_{CO}^{consumed}=0.1434molCO*\frac{3molCO}{1molFe_2O_3} =0.4302molCO[/tex]
Thus, since there are more available than consumed moles of carbon monoxide, we conclude that the iron (III) oxide is the limiting reagent, in such a way the leftover of carbon monoxide turns out into:
[tex]m_{CO}^{leftover}=(0.5682molCO-0.4302molCO)*\frac{28gCO}{1molCO} =0.138molCO*\frac{28gCO}{1molCO} \\m_{CO}^{leftover}=3.864gCO[/tex]
Best regards.
