Chapter 10, Problem 8E

Butane, ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ is a common fuel used for heating homes in areas not served by natural gas. The equation for its combustion is ${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}+1{\text{3O}}_2\to {\text{8CO}}_{\text{2}}+{\text{10H}}_{\text{2}}\text{O}$. All parts of this question are related to this reaction.

a) How many grams of butane can be burned by $1.42$ moles of oxygen?

b) If $\text{9}\text{.43}\text{g}$ of oxygen is used in burning butane, how many moles of water result?

c) Calculate the number of grams of carbon dioxide that will be produced by burning $\text{78}\text{.4}\text{g}$ of butane.

d) How many grams of oxygen are used in a reaction that produces $\text{43}\text{.8}\text{g}$ of water?

Interpretation Introduction

(a)

Interpretation:

The grams of butane that can be burned by $1.42$ moles of oxygen are to be predicted.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in an equation which determines the moles of the reactants and products in the reaction.

Answer to Problem 8E

The grams of butane that can be burned by $1.42$ moles of oxygen are $12.69\text{g}$.

Explanation of Solution

The balanced equation for the reaction for the combustion of butane is given below.

${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}+1{\text{3O}}_2\to {\text{8CO}}_{\text{2}}+{\text{10H}}_{\text{2}}\text{O}$

Therefore, $2$ moles of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ react with $\text{13}$ moles of ${\text{O}}_{\text{2}}$.

Therefore mole to mole ratio are given below.

$2\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}=13\text{mol}{\text{O}}_{\text{2}}$

Therefore, two conversion factors from the mole-to-mole ratio are given below.

$\frac{2\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}{13\text{mol}{\text{O}}_{\text{2}}}\text{and}\frac{13\text{mol}{\text{O}}_{\text{2}}}{2\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}$

The conversion factor to obtain moles of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ from ${\text{O}}_{\text{2}}$ are given below.

$\frac{2\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}{13\text{mol}{\text{O}}_{\text{2}}}$

The molar mass of carbon is $12.01\text{g}{\text{mol}}^{-1}$.

The molar mass of hydrogen is $1.008\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(4\times 12.01\text{g}{\text{mol}}^{-1}\right)+\left(10\times 1.008\text{g}{\text{mol}}^{-1}\right)\\ & =& \text{58}\text{.12}\text{g}{\text{mol}}^{-1}\end{array}$

The conversion factor to determine the grams of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ from the moles ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ of is given below.

$\frac{\text{58}\text{.12}\text{g}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}{1\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}$

The formula to calculate the grams of butane from moles of ${\text{O}}_{\text{2}}$ is given below.

$\text{Grams}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}=\left(\begin{array}{l}\text{Given}\text{moles}\text{of}{\text{O}}_{\text{2}}\times \text{Conversion}\text{factor}\text{to}\text{obtain moles of}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\text{from}{\text{O}}_{\text{2}} \\ \times \text{Conversion}\text{factor}\text{to}\text{obtain grams}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}} \end{array}\right)$ The moles of ${\text{O}}_{\text{2}}$ are $1.42\text{mol}$.

Substitute the mass of ${\text{O}}_{\text{2}}$ and the conversion factors in the above equation.

$\begin{array}{rll}\text{Grams}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}& =& 1.42\text{mol}{\text{O}}_{\text{2}}\times \frac{2\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}{13\text{mol}{\text{O}}_{\text{2}}}\times \frac{\text{58}\text{.12}\text{g}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}{1\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}\\ & =& 12.69\text{g}\end{array}$

Therefore, the moles of ${\text{NH}}_{\text{3}}$ that can be oxidized by $\text{268}\text{g}$ of oxygen are $6.70\text{mol}$.

Conclusion

The grams of butane that can be burned by $1.42$ moles of oxygen are $12.69\text{g}$.

Interpretation Introduction

(b)

Interpretation:

The moles of water that will be produced if $\text{9}\text{.43}\text{g}$ of oxygen is used in burning butane are to be stated.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in an equation which determines the moles of the reactants and products in the reaction.

Answer to Problem 8E

The moles of water that will be produced if $\text{9}\text{.43}\text{g}$ of oxygen is used in burning butane are $\text{0}\text{.2267}\text{mol}$.

Explanation of Solution

The balanced equation for the reaction for the combustion of butane is given below.

${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}+1{\text{3O}}_2\to {\text{8CO}}_{\text{2}}+{\text{10H}}_{\text{2}}\text{O}$

Therefore, $13$ moles of ${\text{O}}_2$ react to give $\text{10}$ moles of ${\text{H}}_{\text{2}}\text{O}$.

Therefore mole to mole ratio are given below.

$13\text{mol}{\text{O}}_2=10\text{mol}{\text{H}}_{\text{2}}\text{O}$

Therefore, two conversion factors from the mole-to-mole ratio are given below.

$\frac{13\text{mol}{\text{O}}_2}{10\text{mol}{\text{H}}_{\text{2}}\text{O}}\text{and}\frac{10\text{mol}{\text{H}}_{\text{2}}\text{O}}{13\text{mol}{\text{O}}_2}$

The conversion factor to obtain moles of ${\text{H}}_{\text{2}}\text{O}$ from ${\text{O}}_2$ is given below.

$\frac{10\text{mol}{\text{H}}_{\text{2}}\text{O}}{13\text{mol}{\text{O}}_2}$

The molar mass of oxygen is $\text{16}\text{.00}\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{O}}_{\text{2}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 2\times 16.00\text{g}{\text{mol}}^{-1}\\ & =& \text{32}\text{.00}\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the conversion factor to determine moles of ${\text{O}}_{\text{2}}$ from grams of ${\text{O}}_{\text{2}}$ is given below.

$\frac{1\text{mol}{\text{O}}_{\text{2}}}{\text{32}\text{.00}\text{g}{\text{O}}_{\text{2}}}$

The formula to calculate the grams of ${\text{H}}_{\text{2}}\text{O}$ from moles of ${\text{O}}_2$ are given below.

$\text{Grams}\text{of}{\text{H}}_2\text{O}=\left(\begin{array}{l}\text{Given}\text{grams}\text{of}{\text{O}}_2\times \text{Conversion}\text{factor}\text{to obtain moles of}{\text{O}}_2 \times \\ \text{Conversion}\text{factor}\text{to obtain}\text{grams}\text{of}{\text{H}}_{\text{2}}\text{O}\end{array}\right)$

The grams of ${\text{O}}_2$ are $\text{9}\text{.43}\text{g}$.

Substitute the grams of ${\text{O}}_2$ and the conversion factors in the above equation.

$\begin{array}{rll}\text{Moles}\text{of}{\text{H}}_{\text{2}}\text{O}& =& \text{9}\text{.43}\text{g}{\text{O}}_2\times \frac{10\text{mol}{\text{H}}_{\text{2}}\text{O}}{13\text{mol}{\text{O}}_2}\times \frac{1\text{mol}{\text{O}}_{\text{2}}}{\text{32}\text{.00}\text{g}{\text{O}}_{\text{2}}}\\ & =& \text{0}\text{.2267}\text{mol}\end{array}$

The moles of water that will be produced if $\text{9}\text{.43}\text{g}$ of oxygen is used in burning butane are $\text{0}\text{.2267}\text{mol}$.

Conclusion

The moles of water that will be produced if $\text{9}\text{.43}\text{g}$ of oxygen is used in burning butane are $\text{0}\text{.2267}\text{mol}$.

Interpretation Introduction

(c)

Interpretation:

The number of grams of carbon dioxide that will be produced by burning $\text{78}\text{.4}\text{g}$ of butane is to be stated.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in an equation which determines the moles of the reactants and products in the reaction.

Answer to Problem 8E

The number of grams of carbon dioxide that will be produced by burning $\text{78}\text{.4}\text{g}$ of butane is $\text{237}\text{.46}\text{g}$.

Explanation of Solution

The balanced equation for the reaction for the combustion of butane is given below.

${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}+1{\text{3O}}_2\to {\text{8CO}}_{\text{2}}+{\text{10H}}_{\text{2}}\text{O}$

Therefore, $2$ moles of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ react to give $\text{8}$ moles of ${\text{CO}}_{\text{2}}$.

Therefore mole to mole ratio is given below.

$2\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}=8\text{mol}{\text{CO}}_{\text{2}}$

Therefore, two conversion factors from the mole-to-mole ratio are given below.

$\frac{2\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}{8\text{mol}{\text{CO}}_{\text{2}}}\text{and}\frac{8\text{mol}{\text{CO}}_{\text{2}}}{2\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}$

The conversion factor to obtain moles of ${\text{CO}}_{\text{2}}$ from ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ is given below.

$\frac{8\text{mol}{\text{CO}}_{\text{2}}}{2\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}$

The molar mass of carbon is $12.01\text{g}{\text{mol}}^{-1}$.

The molar mass of hydrogen is $1.008\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(4\times 12.01\text{g}{\text{mol}}^{-1}\right)+\left(10\times 1.008\text{g}{\text{mol}}^{-1}\right)\\ & =& \text{58}\text{.12}\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the conversion factor to obtain moles of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ from grams of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ is given below.

$\frac{1\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}{\text{58}\text{.12}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}$

The molar mass of carbon is $12.01\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $\text{16}\text{.00}\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{CO}}_{\text{2}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 12.01\text{g}{\text{mol}}^{-1}+\left(2\times 16.00\text{g}\text{mo}\right){\text{l}}^{-1}\\ & =& \text{44}\text{.01}\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the conversion factor to obtain grams of ${\text{CO}}_{\text{2}}$ from moles of ${\text{CO}}_{\text{2}}$ is given below.

$\frac{\text{44}\text{.01}\text{g}{\text{CO}}_{\text{2}}}{1\text{mol}{\text{CO}}_{\text{2}}}$

The formula to calculate the grams ${\text{CO}}_{\text{2}}$ from the grams of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ is given below.

$\text{Grams}\text{of}{\text{CO}}_{\text{2}}=\left(\begin{array}{l}\text{Given}\text{grams}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\times \text{Conversion}\text{factor}\text{to obtain moles of}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\\ \times \text{Conversion}\text{factor}\text{to obtain grams}\text{of}{\text{CO}}_{\text{2}}\\ \times \text{Conversion}\text{factor}\text{to obtain}\text{moles}\text{of}{\text{CO}}_{\text{2}}\end{array}\right)$

The grams of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ is $\text{78}\text{.4}\text{g}$.

Substitute the grams of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ and the conversion factors in the above equation.

$\begin{array}{rll}\text{Grams}\text{of}{\text{CO}}_{\text{2}}& =& \text{78}\text{.4}\text{g}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\times \frac{1\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}{\text{58}\text{.12}\text{g}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}\times \frac{\text{44}\text{.01}\text{g}{\text{CO}}_{\text{2}}}{1\text{mol}{\text{CO}}_{\text{2}}}\times \frac{8\text{mol}{\text{CO}}_{\text{2}}}{2\text{mol}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}\\ & =& \text{237}\text{.46}\text{g}\end{array}$

The number of grams of carbon dioxide that will be produced by burning $\text{78}\text{.4}\text{g}$ of butane is $\text{237}\text{.46}\text{g}$.

Conclusion

The number of grams of carbon dioxide that will be produced by burning $\text{78}\text{.4}\text{g}$ of butane is $\text{237}\text{.46}\text{g}$.

Interpretation Introduction

(d)

Interpretation:

The grams of oxygen that are used in a reaction that produces $\text{43}\text{.8}\text{g}$ of water is to be stated.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in an equation which determines the moles of the reactants and products in the reaction.

Answer to Problem 8E

The grams of oxygen that are used in a reaction that produces $\text{43}\text{.8}\text{g}$ of water is $\text{101}\text{.14}\text{g}$.

Explanation of Solution

The balanced equation for the reaction for the combustion of butane is given below.

${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}+1{\text{3O}}_2\to {\text{8CO}}_{\text{2}}+{\text{10H}}_{\text{2}}\text{O}$

Therefore, $13$ moles of ${\text{O}}_2$ react with $\text{10}$ moles of ${\text{H}}_{\text{2}}\text{O}$.

Therefore mole to mole ratio are given below.

$13\text{mol}{\text{O}}_2=10\text{mol}{\text{H}}_{\text{2}}\text{O}$

Therefore, two conversion factors from the mole-to-mole ratio are given below.

$\frac{13\text{mol}{\text{O}}_2}{10\text{mol}{\text{H}}_{\text{2}}\text{O}}\text{and}\frac{10\text{mol}{\text{H}}_{\text{2}}\text{O}}{13\text{mol}{\text{O}}_2}$

The conversion factor to obtain moles of ${\text{O}}_2$ from ${\text{H}}_{\text{2}}\text{O}$ are given below.

$\frac{13\text{mol}{\text{O}}_2}{10\text{mol}{\text{H}}_{\text{2}}\text{O}}$

The molar mass of oxygen is $\text{16}\text{.00}\text{g}{\text{mol}}^{-1}$.

The molar mass of hydrogen is $\text{1}\text{.008}\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{H}}_{\text{2}}\text{O}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(2\times \text{1}\text{.008}\text{g}{\text{mol}}^{-1}\right)+\text{16}\text{.00}\text{g}{\text{mol}}^{-1}\\ & =& \text{18}\text{.016}\text{g}{\text{mol}}^{-1}\end{array}$

The conversion factor to calculate the moles of ${\text{H}}_{\text{2}}\text{O}$ from grams of ${\text{H}}_{\text{2}}\text{O}$ is given below.

$\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{18}\text{.016}\text{g}{\text{H}}_{\text{2}}\text{O}}$

The molar mass of oxygen is $\text{16}\text{.00}\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{O}}_{\text{2}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 2\times 16.00\text{g}{\text{mol}}^{-1}\\ & =& \text{32}\text{.00}\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the conversion factor to determine grams of ${\text{O}}_{\text{2}}$ from moles of ${\text{O}}_{\text{2}}$ is given below.

$\frac{\text{32}\text{.00}\text{g}{\text{O}}_{\text{2}}}{1\text{mol}{\text{O}}_{\text{2}}}$

The formula to calculate the grams of ${\text{O}}_{\text{2}}$ from grams of ${\text{H}}_{\text{2}}\text{O}$ is given below.

$\text{Grams}\text{of}{\text{O}}_{\text{2}}=\left(\begin{array}{l}\text{Given}\text{grams}\text{of}{\text{H}}_2\text{O}\times \text{Conversion}\text{factor}\text{to obtain moles of}{\text{H}}_2\text{O}\\ \times \text{Conversion}\text{factor}\text{to obtain}\text{grams}\text{of}{\text{O}}_{\text{2}}\\ \times \text{Conversion}\text{factor}\text{to obtain}\text{moles}\text{of}{\text{O}}_{\text{2}}\end{array}\right)$

The grams of ${\text{H}}_{\text{2}}\text{O}$ are $\text{43}\text{.8}\text{g}$.

Substitute the grams of ${\text{H}}_{\text{2}}\text{O}$ and the conversion factors in equation (4).

$\begin{array}{rll}\text{Grams}\text{of}{\text{O}}_{\text{2}}& =& \text{43}\text{.8}\text{g}{\text{H}}_{\text{2}}\text{O}\times \frac{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{18}\text{.016}\text{g}{\text{H}}_{\text{2}}\text{O}}\times \frac{\text{32}\text{.00}\text{g}{\text{O}}_{\text{2}}}{1\text{mol}{\text{O}}_{\text{2}}}\times \frac{13\text{mol}{\text{O}}_2}{10\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ & =& \text{101}\text{.14}\text{g}\end{array}$

Therefore, the grams of oxygen that are used in a reaction that produces $\text{43}\text{.8}\text{g}$ of water is $\text{101}\text{.14}\text{g}$.

Conclusion

The grams of oxygen that are used in a reaction that produces $\text{43}\text{.8}\text{g}$ of water is $\text{101}\text{.14}\text{g}$.