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

Step-by-step 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*\text{ x moles Fe = 2 moles Fe}*\text{ 1.68 moles Fe}_2O_3 \\ \text{ Isolate x } \\ \text{ x moles Fe = }\frac{2moles\text{ Fe}*1.68\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}*\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 = }(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 }*\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}*\text{ molar mass} \end{gathered}`

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

`\begin{gathered} \text{ Mass = 0.05673}*\text{ 28.01} \\ \text{ Mass = 1.589 grams} \end{gathered}`

Therefore, the mass of CO is 1.589 grams