Chapter 9.3, Problem 5RQ

(a)

Interpretation Introduction

Interpretation:

For the reaction of 50.0 g of iron (II) oxide with 10.0 g of C, limiting reactant should be determined.

Concept Introduction:

The reaction in which the reactant is totally consumed when the reaction is completed is

known as limiting reactant.

Answer to Problem 5RQ

Iron (II) oxide is a limiting reactant.

Explanation of Solution

The given reaction is shown below:

  $\text{2FeO}\left(\text{s}\right)\text{+}\text{C}\left(\text{s}\right)\to \text{2Fe}\left(\text{l}\right)\text{+}{\text{CO}}_{\text{2}}$

Given information:

Mass of FeO = 50 g

Mass of C = 10 g

Molar mass of FeO = 72 gmol-1

Molar mass of C = 12 gmol-1

The calculation of moles is shown below:

  $\begin{array}{c}\text{moles}\text{of}\text{FeO}\text{=}\frac{\text{mass}\text{of}\text{FeO}}{\text{molar}\text{mass}\text{of}\text{FeO}}\\ \text{=}\frac{\text{50}\text{g}}{\text{72}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{0}\text{.69}\text{mol}\\ \text{moles}\text{of}\text{C}\text{=}\frac{\text{mass}\text{of}\text{C}}{\text{molar}\text{mass}\text{of}\text{C}}\\ \text{=}\frac{\text{10}\text{g}}{\text{12}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{0}\text{.83}\text{mol}\end{array}$

According to the given reaction 1 mole of FeOreacts with ½ mole of C. Therefore, 0.69 moles of FeOwould react with 0.69/2 = 0.345 moles of C. But there are 0.69 moles of FeOthat means FeOwill get totally consumed in the reaction and thus, will act as a limiting reactant.

(b)

Interpretation Introduction

Interpretation:

For the reaction of 50.0 g of iron (II) oxide and 10.0 g of C, mass of each product should be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 5RQ

Mass of Fe product = 38.64 g

Mass of CO₂ product = 15.18 g

Explanation of Solution

The given reaction is shown below:

  $\text{2FeO}\left(\text{s}\right)\text{+}\text{C}\left(\text{s}\right)\to \text{2Fe}\left(\text{l}\right)\text{+}{\text{CO}}_{\text{2}}$

Given information:

Mass of FeO = 50 g

Mass of C = 10 g

Molar mass of FeO = 72 gmol-1

Molar mass of C = 12 gmol-1

The calculation of mass is shown below:

Mass of Fe produced = molar mass × moles

  = 56 gmol-1 0.69 mol

  = 38.64 g

Mass of CO₂ produced = molar mass × moles

  = 44 gmol-1× 0.345 mol

  = 15.18 g

(c)

Interpretation Introduction

Interpretation:

For 50.0 g of iron (II) oxide reacting with 10.0 g of C, mass of the leftover reactant should be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 5RQ

Mass of left over reactant is 5.82 g

Explanation of Solution

The given reaction is shown below:

  $\text{2FeO}\left(\text{s}\right)\text{+}\text{C}\left(\text{s}\right)\to \text{2Fe}\left(\text{l}\right)\text{+}{\text{CO}}_{\text{2}}$

Given information:

Mass of FeO = 50 g

Mass of C = 10 g

Molar mass of FeO = 72 gmol-1

Molar mass of C = 12 gmol-1

According to the given reaction 1 mole of FeOreacts with ½ mole of C. Therefore, 0.69 moles of FeOwould react with 0.69/2 = 0.345 moles of C and moles of C present is 0.83 mol.

According to the reaction; 2 mol of FeO forms 2 mol of Fe and 1 mol of CO₂. Therefore; 0.69 moles of FeO will form 0.69 moles of Fe and 0.345 moles of CO₂

But there are 0.69 moles of FeOthat means FeOwill get totally consumed in the reaction and thus, will act as a limiting reactant. So, in the reactant carbon will be leftover.

Left over moles of C = no. of moles of C initially − moles of C reached with FeO

  = 0.83 − 0.345 mol

  = 0.485 mol

The calculation of leftover reactant is shown below:

Mass of left-over C = moles × molar mass

  = 0.485 mol × 12 gmol-1

  = 5.82 g