Question

Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (l)For the case where 188 g of Fe2O3 is treated with 59.5 g of CO calculate the following:(i) Determine the limiting reagent and show the calculations (including number ofmoles) leading to your conclusion.

Answer 1

$Fe_2O_{3(s)}+3CO_{(g)}\rightarrow2Fe_{(s)}+3CO_{2(l)}$

We will firstly convert the mass to moles:

$\begin{gathered} _nFe_2O_{3(g)}=\frac{mass}{molar\text{ }mass} \\ \\ _nFe_2O_{3(g)}=\frac{188.0g}{159.69gmol^{-1}} \\ _nFe_2O_{3(g)}=1.18moles \\ \\ _nCO_{(g)}=\frac{mass}{molar\text{ }mass} \\ \\ _nCO_{(g)}=\frac{59.5g}{28.01gmol^{-1}} \\ _nCO_{(g)}=2.12moles \end{gathered}$

We will now determine the theoretical yield of each reactant to find the limiting reactant. The first reactant we run out of is the limiting reactant.

To use all of the Fe2O3 how many moles of CO do we need:

$\begin{gathered} 1.18molFe_2O_3\times\frac{3mol\text{ }CO}{1molFe2O3}=3.54mol\text{ }CO \\ \end{gathered}$

To use all CO, how many moles of Fe2O3 do we need:

$2.12molCO\times\frac{1mol\text{ }Fe2O3}{3mol\text{ }CO}=0.71\text{ }molFe_2O_3$

We need 3.54 moles of CO but only have 2.12 moles. This indicate that the CO is the limiting reactant.