Chapter 15, Problem 23E

Interpretation Introduction

(a)

Interpretation:

The stoichiometry concept map is to be drawn and the mass of reacted methane $\left({\text{CH}}_{\text{4}}\right)$ in the reaction is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Answer to Problem 23E

The stoichiometry concept map is shown below.

The mass of the reacted methane is $\text{2}\text{.23}\text{g}$.

Explanation of Solution

It is given that $\text{5}\text{.00}\text{g}$ of water is produced after the reaction stated below.

${\text{CH}}_{\text{4}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$…(1)

The stoichiometry concept map is shown below.

Figure 1

The chemical reaction (1) should be balanced first to calculate the mass of reacted methane.

The number of oxygen atoms is not equal on both sides. A coefficient of $2$ is added in front of ${\text{O}}_{\text{2}}$ to balance the chemical reaction.

${\text{CH}}_{\text{4}}\left(g\right)+2{\text{O}}_{\text{2}}\left(g\right)\to {\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The number of hydrogen atoms is not equal on both sides. A coefficient of $2$ is added in front of ${\text{H}}_{\text{2}}\text{O}$ to balance the number of hydrogen atoms.

The balanced equation is shown below.

${\text{CH}}_{\text{4}}\left(g\right)+2{\text{O}}_{\text{2}}\left(g\right)\to {\text{CO}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$…(2)

The mass of methane reacted is equal to the mass of ${\text{H}}_{\text{2}}\text{O}$ produced. The above reaction shows that $\text{1}\text{mole}$ of methane consumed in the reaction to form $\text{2}\text{mol}$ of water.

The molar mass of hydrogen is $1.01\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-1}$.

$\begin{array}{rll}\text{Total}\text{molar}\text{mas}\text{of}{\text{H}}_{\text{2}}\text{O}& =& \left(2\times 1.01\text{g}{\text{mol}}^{-1}\right)+16.00\text{g}{\text{mol}}^{-1}\\ & =& 2.02\text{g}{\text{mol}}^{-1}+16.00\text{g}{\text{mol}}^{-1}\\ & =& 18.02\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the number of moles of methane is calculated from the number of moles of water using a stoichiometric ratio as shown below.

$\begin{array}{l}1\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}=\text{Molar}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}\\ \text{1}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}=\text{18}\text{.02}\text{g}\\ \frac{\text{1}\text{mole}\text{of}{\text{H}}_2\text{O}}{18.02\text{g}}=\text{1}\text{g}\text{of}{\text{H}}_2\text{O}\end{array}$

For $\text{5}\text{g}$ of ${\text{H}}_{\text{2}}\text{O}$, the number of moles of methane is shown below. $\begin{array}{ccc}\text{Number}\text{of}\text{moles}\text{of}{\text{CH}}_4& =& \text{5}\text{.0g}\times \frac{\text{1}\text{mol}\text{of}{\text{H}}_{\text{2}}\text{O}}{\text{18}\text{.02}\text{g}\text{of}{\text{H}}_{\text{2}}\text{O}}\times \frac{\text{1}\text{mol}\text{of}{\text{CH}}_{\text{4}}}{\text{2}\text{mol}\text{of}{\text{H}}_{\text{2}}\text{O}}\\ & =& \text{0}\text{.1387}\text{mol}{\text{CH}}_4\\ \approx \text{0}\text{.139}\text{mol}{\text{CH}}_4\end{array}$

The mass of methane reacted can be calculated from the number of moles of methane which is $\text{0}\text{.139}\text{mol}$

The molar mass of carbon is $\text{12}\text{.01}\text{g}{\text{mol}}^{-1}$.

The molar mass of hydrogen is $1.0\text{1}\text{g}{\text{mol}}^{-1}$.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}\text{of}{\text{CH}}_{\text{4}}& =& \text{Molar}\text{mass}\text{of}\text{C}+\left(\text{4}\times \text{molar}\text{mass}\text{of}\text{H}\right)\\ & =& \text{12}\text{.02}\text{g}{\text{mol}}^{-\text{1}}+\left(4\times 1.01\text{g}{\text{mol}}^{-\text{1}}\right)\\ & =& 12.02\text{g}{\text{mol}}^{-\text{1}}+4.04\text{g}{\text{mol}}^{-\text{1}}\\ & =& 16.06\text{g}{\text{mol}}^{-\text{1}}\end{array}$

The number of moles of methane can be calculated from the relation shown below.

$\begin{array}{l}\text{1}\text{mole}\text{of}{\text{CH}}_4=\text{Molar}\text{mass}\text{of}{\text{CH}}_{\text{4}}\\ \text{1}\text{mole}\text{of}{\text{CH}}_{\text{4}}=16.06\text{g}{\text{mol}}^{\text{-1}}\end{array}$

The mass of methane for $0.\text{139}\text{moles}$ is shown below.

$\begin{array}{rll}\text{Mass}\text{of}{\text{CH}}_{\text{4}}& =& \text{0}\text{.139}\text{mol×}\frac{\text{16}\text{.06}\text{g}\text{}{\text{CH}}_{\text{4}}}{\text{1}\text{mol}{\text{CH}}_{\text{4}}}\\ & =& 2.23\text{g}{\text{CH}}_{\text{4}}\end{array}$

Therefore, the mass of methane is $\text{2}\text{.23}\text{g}$.

Conclusion

The stoichiometry concept map is shown in Figure 1. The mass of methane consumed in the reaction is $\text{2}\text{.23}\text{g}$.

Interpretation Introduction

(b)

Interpretation:

The volume of carbon dioxide $\left({\text{CO}}_{\text{2}}\right)$ produced at $\text{STP}\text{}$ is to be stated.

Concept introduction:

Stoichiometry measures the amount of reactant and product consumed or formed in a chemical reaction. It gives the relationship between the reactants and products. Stoichiometry concept maps are used to determine the relation of reactant and product in terms of stoichiometry.

Answer to Problem 23E

The stoichiometry map is shown below.

The volume of carbon dioxide produced at $\text{STP}$ is $\text{3}\text{.11}\text{L}$.

Explanation of Solution

The stoichiometry concept mass is shown below.

Figure 1

The balanced chemical reaction is stated in part (a) named as equation (2) is used to calculate the volume of ${\text{CO}}_{\text{2}}$ gas.

${\text{CH}}_{\text{4}}\left(g\right)+2{\text{O}}_{\text{2}}\left(g\right)\to {\text{CO}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The equation (2) shows that 1 mole of methane is consumed to form 1mole of carbon dioxide.

$\text{1}\text{mole}\text{of}{\text{CH}}_4=\text{22}\text{.4}\text{L}{\text{CO}}_{\text{2}}$

Therefore, the volume for $\text{0}\text{.139}\text{moles}$ is calculated as shown below.

$\begin{array}{rll}\text{Volume}\text{of}{\text{CO}}_{\text{2}}\text{produced}& =& \text{0}\text{.139}\text{mol}{\text{CH}}_{\text{4}}\times \frac{\text{1}\text{mol}{\text{CO}}_{\text{2}}}{\text{1}\text{mole}{\text{CH}}_{\text{4}}}\times \frac{\text{22}\text{.4}\text{L}{\text{CO}}_{\text{2}}}{\text{1}\text{mole}{\text{CO}}_{\text{2}}}\\ & =& \text{3}\text{.11}\text{L}{\text{CO}}_{\text{2}}\end{array}$

Therefore, the volume of ${\text{CO}}_{\text{2}}$ produced is $\text{3}\text{.11}\text{L}$.

Conclusion

The volume of ${\text{CO}}_{\text{2}}$ gas produced at $\text{STP}$ is $\text{3}\text{.11}\text{L}$.