Chapter 12.1, Problem 4SP

Interpretation Introduction

Interpretation: To interpret the equation in terms of relative numbers of moles, the volume of gas at STP, and masses of reactants and products.

Concept introduction: A mole is a SI unit of the amount of substance. In 1 mole of a substance, there are $6.022\times {10}^{23}$ particles/ions/molecules.

Answer to Problem 4SP

In a given equation, there are ${\text{1 mol C}}_2{\text{H}}_4\text{(}g\text{) }+{\text{ 1 mol O}}_2(g)$ moles of reactant and ${\text{1 mol CO}}_2\text{(}g\text{) }+{\text{ 1 mol H}}_2\text{O}(g)$ moles of product. The volume of gas on the reactant side are $\text{22}{\text{.4 L C}}_2{\text{H}}_4\text{(}g\text{) }+\text{22}{\text{.4 L O}}_2(g)$ and on the products side are $\text{22}{\text{.4 L CO}}_2\text{(}g\text{) }+\text{ 22}{\text{.4 L H}}_2\text{O}(g)$ . The mass of reactants is: $\text{28}\text{.05 }g{\text{ C}}_2{\text{H}}_4+\text{ 16 }g{\text{ O}}_2$ and The mass of products is: $\text{44 }g{\text{ CO}}_2+\text{ 18 }g{\text{ H}}_2\text{O}$

Explanation of Solution

The balanced equation given is:

  ${\text{C}}_2{\text{H}}_4\text{(}g\text{)}+{\text{O}}_2(g\text{)}\to {\text{CO}}_2\text{(}g{\text{)+H}}_{2}O(g)$

On the reactant side: The number of moles of ${\text{C}}_2{\text{H}}_4\text{(}g\text{)}$ and ${\text{O}}_2(g\text{)}$ are:

  ${\text{1 mol C}}_2{\text{H}}_4\text{(}g\text{) }+{\text{ 1 mol O}}_2(g)$

On the product side: The number of moles of ${\text{CO}}_2\text{(}g\text{)}$ and ${\text{H}}_{2}O(g)$ are:

  ${\text{1 mol CO}}_2\text{(}g\text{) }+{\text{ 1 mol H}}_2\text{O}(g)$

Since at STP, 1 mole of any gas has a molar volume equal to 22.4 L, thus,

On the reactant side: The volume of gases at STP are:

  $\text{22}{\text{.4 L C}}_2{\text{H}}_4\text{(}g\text{) }+\text{22}{\text{.4 L O}}_2(g)$

On the product side: The volume of gases at STP are:

  $\text{22}{\text{.4 L CO}}_2\text{(}g\text{) }+\text{ 22}{\text{.4 L H}}_2\text{O}(g)$

Since 1 mole of each species reacted and formed in the given equation thus, in each case mass is equal to the molar mass of the species.

Therefore,

On the reactant side: The mass of reactants are:

  $\text{28}\text{.05 }g{\text{ C}}_2{\text{H}}_4+\text{ 16 }g{\text{ O}}_2$

On the product side: The mass of products are:

  $\text{44 }g{\text{ CO}}_2+\text{ 18 }g{\text{ H}}_2\text{O}$