Chapter 12, Problem 42A

Interpretation Introduction

Interpretation : The balanced equation is to be interpreted in terms of the relative number of moles, volumes of gas at STP, and masses of reactants and products.

Concept Introduction : The coefficients of a balanced chemical equation that are transformed into moles are used to calculate the mole ratio, which is a conversion factor.

Mole ratios are utilized when converting between a certain number of moles of one reactant or product and moles of another reactant or product.

Answer to Problem 42A

The balanced chemical equation is given as

  ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\left(g\right)+8{\text{O}}_{\text{2}}\left(g\right) \to 5{\text{CO}}_{\text{2}}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The volume of gas at STP is given as

  $\text{22}\text{.4}\text{L}{\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\left(g\right)+179.2\text{L}{\text{O}}_{\text{2}}\left(g\right) \to 112{\text{L CO}}_{\text{2}}\left(g\right)+134.4\text{L}\text{}{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The mass of the reactants is $\text{328}\text{.146}\text{g/mol}$ .

The mass of the products is $\text{328}\text{.146}\text{g/mol}$

Explanation of Solution

The given chemical equation is as follows:

  ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right) \to {\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The balanced chemical equation is given as

  ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\left(g\right)+8{\text{O}}_{\text{2}}\left(g\right) \to 5{\text{CO}}_{\text{2}}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The moles of each reacted in the reaction is given as

  $\text{1}{\text{mole C}}_{\text{5}}{\text{H}}_{\text{12}}\left(g\right)+8{\text{mole O}}_{\text{2}}\left(g\right) \to 5{\text{mole CO}}_{\text{2}}\left(g\right)+6{\text{mole H}}_{\text{2}}\text{O}\left(g\right)$

At STP, 1 mole is equal to $22.4\text{L}$ .

The volume of gas at STP in the chemical equation is given as

  $\text{22}\text{.4}{\text{L C}}_{\text{5}}{\text{H}}_{\text{12}}\left(g\right)+\left(8\times 22.4\text{L}\right){\text{mole O}}_{\text{2}}\left(g\right) \to \left(5\times 22.4\text{L}\right){\text{mole CO}}_{\text{2}}\left(g\right)+\left(6\times 22.4\text{L}\right){\text{mole H}}_{\text{2}}\text{O}\left(g\right)$

Further, multiplying the values, the volume of gas at STP is given as

  $\text{22}\text{.4}\text{L}{\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\left(g\right)+179.2\text{L}{\text{O}}_{\text{2}}\left(g\right) \to 112{\text{L CO}}_{\text{2}}\left(g\right)+134.4\text{L}\text{}{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Molar mass of $\text{C}$ is $12.01\text{g/mol}$ .

Molar mass of O is $\text{16}\text{.00}\text{g/mol}$ .

Molar mass of $\text{H}$ is $\text{1}\text{.008}\text{g/mol}$ .

The molar mass of the reactants and products is to be calculated.

The reactants are ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}$ and ${\text{8O}}_{\text{2}}$ and the products are ${\text{5CO}}_{\text{2}}$ and ${\text{6H}}_{\text{2}}\text{O}$ .

  $\begin{array}{c}\text{Mass of}{\text{C}}_{\text{5}}{\text{H}}_{\text{12}}=5\text{M}\left(\text{C}\right)\text{+12M}\left(\text{H}\right)\\ =\left(\left(5\times 12.01\right)+\left(12\times 1.008\right)\right)\text{g/mol}\\ \text{=}\text{72}\text{.146}\text{g/mol}\end{array}$

Mass of ${\text{O}}_2$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{O}}_2=2\text{M}\left(\text{O}\right)\\ =\left(2\times 16.00\right)\text{g/mol}\\ \text{=32}\text{.00}\text{g/mol}\end{array}$

Mass of ${\text{CO}}_2$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{CO}}_2=\text{M}\left(\text{C}\right)\text{+2M}\left(\text{O}\right)\\ =\left(12.01+\left(2\times 16.00\right)\right)\text{g/mol}\\ \text{= 44}\text{.01}\text{g/mol}\end{array}$

Mass of ${\text{H}}_2\text{O}$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{H}}_2\text{O}=2\text{M}\left(\text{H}\right)\text{+M}\left(\text{O}\right)\\ =\left(\left(2\times 1.008\right)+16.00\right)\text{g/mol}\\ \text{= 18}\text{.016}\text{g/mol}\end{array}$

The mass of the reactant is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{C}}_{\text{5}}{\text{H}}_{\text{12}}+8{\text{O}}_{\text{2}}=\text{M}\left({\text{C}}_5{\text{H}}_{12}\right)+\text{M}\left({\text{O}}_2\right)\\ =\left(\text{72}\text{.146+}\left(8\times 16.00\right)\right)\text{g/mol}\\ \text{= 328}\text{.146}\text{g/mol}\end{array}$

The mass of the products is given as follows:

  $\begin{array}{c}\text{Mass of}5{\text{CO}}_{\text{2}}+6{\text{H}}_{\text{2}}\text{O}=5\text{M}\left({\text{CO}}_2\right)+6\text{M}\left({\text{H}}_2\text{O}\right)\\ =\left(\left(5\times 44.\text{01}\right)+\left(6\times 18.016\right)\right)\text{g/mol}\\ \text{= 328}\text{.146}\text{g/mol}\end{array}$