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

`m_(CO)^(leftover)=3.864gCO`

Step-by-step 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*(1molFe_2O_3)/(160gFe_2O_3)=0.1434molFe_2O_3\\ n_(CO)=15.91molCO*(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*(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)*(28gCO)/(1molCO) =0.138molCO*(28gCO)/(1molCO) \\m_(CO)^(leftover)=3.864gCO`

Best regards.