Question

Iron(III) oxide reacts with carbon monoxide according to the equation: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) A reaction mixture initially contains 22.95 g Fe2O3 and 15.91 g CO. Assume that the reaction will progress to 100% completion. What mass (in g) of the excess reactant is leftover?

Answer 1

The answer is B because i just took the test

Answer 2

Answer:

$m_{CO}^{leftover}=3.864gCO$

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:

$n_{Fe_2O_3}=22.95gFe_2O_3*\frac{1molFe_2O_3}{160gFe_2O_3}=0.1434molFe_2O_3\\ n_{CO}=15.91molCO*\frac{1molCO}{28gCO}=0.5682molCO$

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:

$n_{CO}^{consumed}=0.1434molCO*\frac{3molCO}{1molFe_2O_3} =0.4302molCO$

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:

$m_{CO}^{leftover}=(0.5682molCO-0.4302molCO)*\frac{28gCO}{1molCO} =0.138molCO*\frac{28gCO}{1molCO} \\m_{CO}^{leftover}=3.864gCO$

Best regards.