Question

сооAn alternative method for the preparation of pure iron (Fe) is as follows from the followingunbalanced equationFO)сооFe) -(a) Balance the equation(b) Calculate the number of moles of iron that can be produced from 1.66 moles of 20:0(c) Calculate the amount (in grams) of CO needed to react with 3.02 g of FeO)(d) Calculate the amount (in grams) of iron (Fe) can be produced from 100 g of Fe2O, and6.00 g of CO(©) Calculate the amount of excess reagent that remains(1) Calculate the % yield if 6.75 g of iron (Fe) is actually formed

Answer 1

ANSWER

EXPLANATION

Given that;

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

To answer the question below, follow the steps below

Part A

Step 1: Balance the chemical equation of the reaction

The balanced equation of the above reaction is given below as

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

Part B

Calculate the number of moles of iron that can be produced from 1.68 moles of Fe2O3

In the above reaction, 1 mole of Fe2O3 reacts with 3 moles of CO to give 2 moles of iron and 3 moles of CO2

The number of moles of Fe can be determined by using a stoichiometry ratio

Let the number of moles of Fe be x

$\begin{gathered} \text{ 1 mole Fe}_2O_3\text{ }\rightarrow\text{ 2 moles Fe} \\ \text{ 1.68 moles Fe}_2O_3\text{ }\rightarrow\text{ x moles Fe} \\ \text{ Cross multiply} \\ \text{ 1 mole Fe}_2O_3\times\text{ x moles Fe = 2 moles Fe}\times\text{ 1.68 moles Fe}_2O_3 \\ \text{ Isolate x } \\ \text{ x moles Fe = }\frac{2moles\text{ Fe}\times1.68\cancel{molesFe_2}O_3}{1\cancel{moleFe_2}O_3} \\ \text{ x moles Fe = 2 moles Fe x 1.68 moles} \\ x\cancel{molesFe}=2\cancel{molesFe}\times\text{ 1.68} \\ \text{ x = 2 x 1.68} \\ \text{ x = 3.36 moles} \end{gathered}$

Therefore, the number of moles of Fe is 3.36 moles

Part C

Calculate the amount (in grams) of CO needed to react with 3.02 grams of Fe2O3

Step 1: Find the number of moles of Fe2O3 using the below formula

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

Recall, that the molar mass of Fe2O3 is 159.69 g/mol, and the mass of Fe2O3 is 3.02 grams

$\begin{gathered} \text{ Mole = }\frac{3.02}{159.69} \\ \text{ Mole = 0.01891} \end{gathered}$

Since the number of moles of Fe2O3 is 0.01891, hence, we can find the number of moles of CO using a stoichiometry ratio

In the above reaction, 1 mole of Fe2O3 will react with 3 moles of CO

$\begin{gathered} \text{ If 1 mole of Fe}_2O_3\text{ gives 3 moles of CO} \\ \text{ Then, the number of moles of CO can be calculated below as} \\ \text{ Moles of CO = 3 }\times\text{ 0.01891} \\ \text{ Moles of CO = 0.05673 mole} \end{gathered}$

The next step is to find the amount of CO in grams using the below formula

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

Recall, that the molar mass of CO is 28.01 g/mol

$\begin{gathered} \text{ Mass = 0.05673}\times\text{ 28.01} \\ \text{ Mass = 1.589 grams} \end{gathered}$

Therefore, the mass of CO is 1.589 grams