Chapter 9, Problem 54A

Interpretation Introduction

Interpretation: The mole ratio from which the number of moles of oxygen required for given moles of ${\text{C}}_3{\text{H}}_8(g)$ can be calculated needs to be determined. Also, the mole ratio from which the number of moles of each product formed from given moles of ${\text{C}}_3{\text{H}}_8(g)$ can be calculated needs to be determined.

Concept introduction:The ratio of mass of substance to its molar mass is said to be number of moles of that substance.

The ratio of amount of any two compounds involved in a chemical reaction in terms of mole is said to be mole ratio, and used as a conversion factor between reactants and products.

Explanation of Solution

The complete balanced equation is:

  ${\text{C}}_3{\text{H}}_8\left(\text{g}\right){\text{ + 5O}}_{\text{2}}\left(\text{g}\right)\to 3{\text{CO}}_{\text{2}}\left(\text{g}\right){\text{ + 4H}}_{\text{2}}\text{O}\left(\text{g}\right)$

• From the balanced chemical reaction, the mole ratio of ${\text{C}}_3{\text{H}}_8(g)$ and oxygen is 1:5 that means 1 mole of ${\text{C}}_3{\text{H}}_8(g)$ reacts with 5 moles of oxygen. So, the mole ratio that enables to calculate the number of moles of oxygen required to react with given number of moles of ${\text{C}}_3{\text{H}}_8(g)$ is:

  ${\text{Given mol of C}}_3{\text{H}}_8(g)\text{×}\frac{{\text{5 mol of O}}_{\text{2}}}{{\text{1 mol of C}}_3{\text{H}}_8(g)}{\text{ = Required mol of O}}_{\text{2}}$

For example: if the given mole of ${\text{C}}_3{\text{H}}_8(g)$ is 2 moles, then required moles of oxygen will be:

  ${\text{2 of C}}_3{\text{H}}_8(g)\text{×}\frac{{\text{5 mol of O}}_{\text{2}}}{{\text{1 mol of C}}_3{\text{H}}_8(g)}{\text{ = 10 mol of O}}_{\text{2}}$

• From the balanced chemical reaction, the mole ratio of ${\text{C}}_3{\text{H}}_8(g)$ and ${\text{CO}}_{\text{2}}$ is 1:3 that means 1 mole of ${\text{C}}_3{\text{H}}_8(g)$ reacts with 3 moles of ${\text{CO}}_{\text{2}}$ . So, the mole ratio that enables to calculate the number of moles of ${\text{CO}}_{\text{2}}$ required to produce from the given number of moles of ${\text{C}}_3{\text{H}}_8(g)$ is:

  ${\text{Given mol of C}}_3{\text{H}}_8(g)\text{×}\frac{{\text{3 mol of CO}}_{\text{2}}}{{\text{1 mol of C}}_3{\text{H}}_8(g)}{\text{ = Required mol of CO}}_{\text{2}}$

For example: if the given mole of ${\text{C}}_3{\text{H}}_8(g)$ is 2 moles then, required moles of ${\text{CO}}_{\text{2}}$ will be:

  ${\text{2 of C}}_3{\text{H}}_8(g)\text{×}\frac{{\text{3 mol of CO}}_{\text{2}}}{{\text{1 mol of C}}_3{\text{H}}_8(g)}{\text{ = 6 mol of CO}}_{\text{2}}$

• From the balanced chemical reaction, the mole ratio of ${\text{C}}_3{\text{H}}_8(g)$ and ${\text{H}}_{\text{2}}\text{O}$ is 1:4 that means 1 mole of ${\text{C}}_3{\text{H}}_8(g)$ reacts with 4 moles of ${\text{H}}_{\text{2}}\text{O}$ . So, the mole ratio that enables to calculate the number of moles of ${\text{H}}_{\text{2}}\text{O}$ required to produce from the given number of moles of ${\text{C}}_3{\text{H}}_8(g)$ is:

  ${\text{Given mol of C}}_3{\text{H}}_8(g)\text{×}\frac{{\text{4 mol of H}}_{\text{2}}\text{O}}{{\text{1 mol of C}}_3{\text{H}}_8(g)}{\text{ = Required mol of H}}_{\text{2}}\text{O}$

For example: if the given mole of ${\text{C}}_3{\text{H}}_8(g)$ is 2 moles then required moles of ${\text{H}}_{\text{2}}\text{O}$ will be:

  ${\text{2 of C}}_3{\text{H}}_8(g)\text{×}\frac{{\text{4 mol of H}}_{\text{2}}\text{O}}{{\text{1 mol of C}}_3{\text{H}}_8(g)}{\text{ = 8 mol of H}}_{\text{2}}\text{O}$