Question

The reaction of 192g of Fe2O3 with 82.5g of CO produces 72.9 of Fe Fe2O3 + 3CO => 2Fe + 3CO2 Calculate the theoretical yield of iron in grams.

Answer 1

ANSWER

The theoretical yield of Fe is 109.64 grams

EXPLANATION

Given that;

The mass of Fe2O3 is 192 grams

The mass of CO is 82.5 grams

Follow the steps below to find the theoretical yield of Iron

Steps 1; Find the number of moles of Fe2O3 and CO using the below formula

$\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}$

Recall, that the molar mass of Fe2O3 and CO are 159.69 g/mol and 28.01 g/mol

$\begin{gathered} \text{ For Fe}_2O_3 \\ \text{ mole = }\frac{\text{ 192}}{\text{ 159.69}} \\ \text{ mole = 1.202 moles} \\ \\ \text{ For CO} \\ \text{ mole = }\frac{\text{ 82.5}}{\text{ 28.01}} \\ \text{ mole = 2.945 moles} \end{gathered}$

Step 2; Determine the limiting raectant of the reaction

To determine the limiting reactant, divide the number of moles by the co-efficient of each reactant

$\begin{gathered} \text{ For Fe}_2O_3 \\ \text{ mol/wt of Fe}_2O_3\text{ = }\frac{\text{ 1.201 }}{\text{ 1}} \\ \text{ mol/wt of Fe}_2O_3\text{ = 1.202 mol} \\ \\ \text{ for CO} \\ \text{ mol/wt of CO = }\frac{\text{ 2.945}}{\text{ 3}} \\ \text{ mol/wt of CO = 0.982 mol} \end{gathered}$

From the calculations, the limiting reactant of the reaction is CO

Step 3; Find the number of moles of Fe using a stoichiometry ratio

Let x represents the number of moles of Fe

$\begin{gathered} \text{ 3 moles CO }\rightarrow\text{ 2 moles Fe} \\ \text{ 2.945 moles CO }\rightarrow\text{ x moles Fe} \\ \text{ cross multiply} \\ \text{ 3 moles CO}\times\text{ x moles Fe = 2 moles Fe }\times\text{ 2.945 moles CO} \\ \text{ x moles Fe = }\frac{\text{ 2 moles Fe}\times2.945mole\cancel{CO}}{3moles\cancel{CO}} \\ \\ \text{ x mole Fe = }\frac{\text{ 2 }\times\text{ 2.945}}{\text{ 3}} \\ \\ \text{ x mole Fe = }\frac{\text{ 5.89}}{\text{ 3}} \\ \text{ x mole Fe = 1.963} \end{gathered}$

The number of moles of Fe is 1.963 moles

Step 4; Find the theoretical yield of Fe

$\begin{gathered} \text{ mole = }\frac{\text{ mass}}{\text{ molar mass}} \\ \text{ cross multiply} \\ \text{ mass = mole }\times\text{ molar mass} \end{gathered}$

Recall, molar mass of Fe is 55.845 g/mol

$\begin{gathered} \text{ mass = 1.963 }\times\text{ 55.845} \\ \text{ mass = 109.64 grams} \end{gathered}$

Therefore, the theoretical yield of Fe is 109.64 grams