Question

Mining companies can obtain iron from iron ore by treatment with carbon monoxide asdescribed in the equation below.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:What mass of solid iron (Fe) forms when the reaction goes to completion?

Answer 1

ANSWER

The mass of solid iron formed is 78.93 grams

EXPLANATION

Given that;

The mass of Fe2O3 is 188g

The mass of CO reacted is 59.5g

Follow the steps below to find the mass of Fe reacted

Step 1; Write the balanced equation for the reaction

$\text{ Fe}_2O_{3(s)}\text{ + 3CO}_{(g)}\text{ }\rightarrow\text{ 2Fe}_{(s)}\text{ + 3CO}_{2(g)}$

Step 2; Find the number of moles of each reactant

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

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

$\begin{gathered} \text{ For Fe}_2O_3 \\ \text{ mole = }\frac{\text{ mass}}{\text{ molar mass}} \\ \\ \text{ mole = }\frac{\text{ 188}}{\text{ 159.69}} \\ \text{ mole = 1.18 moles} \\ \\ \text{ For CO} \\ \text{ mole = }\frac{59.5}{\text{ 28.01}} \\ \text{ mole = 2.12 moles} \end{gathered}$

The mole of Fe2O3 is 1.18 moles and the mole of CO is 2.12 moles

Hence, the limitig reagent of the reaction is CO

Step 3; Find the mole of Fe using stoichiometry ratio

Let x represents the number of moles of Fe

$\begin{gathered} \text{ 3 moles CO }\rightarrow\text{ 2 moles Fe} \\ \text{ 2.12 moles CO }\rightarrow\text{ x moles Fe} \\ \text{ cross multiply} \\ \text{ 3 moles CO }\times\text{ x moles Fe = 2 moles Fe }\times\text{ 2.12 moles CO} \\ \text{ Isolate x moles Fe} \\ \text{ x moles Fe = }\frac{2\text{ moles Fe }\times2.12moles\cancel{CO}}{3moles\cancel{CO}} \\ \\ \text{ x moles Fe = }\frac{2\text{ }\times\text{ 2.12}}{3} \\ \\ \text{ x moles Fe = }\frac{4.24}{\text{ 3}} \\ \text{ x moles Fe = 1.41 moles} \end{gathered}$

Therefore, the number of moles of Fe is 1.41 moles

Step 4; Find the mass of Fe

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

The molar mass of Fe is 55.845u

$\begin{gathered} \text{ mass = 1.41 }\times\text{ 55.845} \\ \text{ mass = 78.93 grams} \end{gathered}$

Therefore, the mass of solid iron formed is 78.93 grams