Chapter 3.4, Problem 3.20BFP

Interpretation Introduction

Interpretation:

The mass of carbon dioxide produced when $4.65\text{ g}$ of butane is burned in $10.0\text{ g}$ of oxygen is to be calculated. Also, the mass of excess reactant that remains unreacted when the reaction is over is to be calculated.

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass. Also, the amounts of substances in a balanced chemical reaction are stoichiometrically equivalent to each other.

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction.

Amount(mol) of excess reactant is reactant left after the formation of maximum amount(mol) of products.

The formula to calculate moles is as follows:

  $\text{Amount}\left(\text{mol}\right)=\frac{\text{mass}}{\text{molar mass}}$        (1)

Answer to Problem 3.20BFP

The mass of carbon dioxide produced is $8.46\text{ g}$ and the mass of excess reactant that remains unreacted is $1.86\text{g}$.

Explanation of Solution

The balanced equation for the combustion of butane is as follows:

  ${\text{2C}}_4{\text{H}}_{10}\left(g\right)+13{\text{O}}_2\left(s\right)\to 8{\text{CO}}_2\left(s\right)+10{\text{H}}_{\text{2}}\text{O}$

Substitute $4.65\text{ g}$ for mass and $58.12\text{ g/mol}$ for molar mass in equation (1) to calculate amount (mol) of ${\text{C}}_4{\text{H}}_{10}$.

  $\begin{array}{c}{\text{Moles of C}}_4{\text{H}}_{10}=\frac{4.65\text{ g}}{58.12\text{ g/mol}}\\ =0.080006\text{ mol}\end{array}$

Substitute $10\text{ g}$ for mass and $32.00\text{ g/mol}$ for molar mass mass in equation (1) to calculate amount (mol) of ${\text{O}}_2$.

  $\begin{array}{c}{\text{Moles of O}}_2=\frac{10\text{ g}}{32.00\text{ g/mol}}\\ =0.3125\text{ mol}\end{array}$

According to the balanced chemical equation, the stoichiometric ratio between ${\text{C}}_4{\text{H}}_{10}$ and ${\text{CO}}_2$ is $2:8$. Therefore, the number of moles of ${\text{CO}}_2$ is calculated as:

  $\begin{array}{c}{\text{Amount of CO}}_2=\left(0.080006\text{ mol}\right)\left(\frac{{\text{8 mol CO}}_2}{2{\text{ mol C}}_4{\text{H}}_{10}}\right)\\ =0.640055\text{ mol}\end{array}$

According to the balanced chemical equation, the stoichiometric ratio between ${\text{O}}_2$ and ${\text{CO}}_2$ is $13:8$. Therefore, the number of moles of ${\text{CO}}_2$ is calculated as:

  $\begin{array}{c}{\text{Amount of CO}}_2=\left(0.3125\text{ mol}\right)\left(\frac{{\text{8 mol O}}_2}{{\text{13 mol CO}}_2}\right)\\ =0.192307\text{ mol}\end{array}$

${\text{O}}_2$ is the limiting reactant, as it produces less amount of product ${\text{CO}}_2$.

The formula to calculate the mass is as follows:

  $\text{Mass}=\left(\text{amount(mol)}\right)\left(\text{molecular mass}\right)$        (2)

Substitute $0.192307\text{ mol}$ for amount(mol) and $44.01\text{ g/mol}$ for molecular mass in equation (2) to calculate mass of ${\text{CO}}_2$.

  $\begin{array}{c}{\text{Mass of CO}}_2=\left(0.192307\text{ mol}\right)\left(44.01\text{ g/mol}\right)\\ =8.4635\text{ g}\\ \approx 8.46\text{ g}\end{array}$

The remaining mass of the excess reagent can be calculated from the amount of ${\text{CO}}_2$ combining with the limiting reagent.

  $\begin{array}{c}{\text{Moles of C}}_4{\text{H}}_{10}\text{ required}=\left[\begin{array}{l}\left(0.3125\text{ mol}\right)\\ \left(\frac{\text{2 mol}{\text{C}}_4{\text{H}}_{10}}{\text{13 mol}\text{Al}}\right)\end{array}\right]\\ =0.0480769\text{ mol}\end{array}$

Substitute $0.0480769\text{ mol}$ for amount (mol) and $58.12\text{ g/mol}$ for molecular mass in equation (2) to calculate the excess amount (mol) of ${\text{C}}_4{\text{H}}_{10}$.

  $\begin{array}{c}{\text{Mass of C}}_4{\text{H}}_{10}=\left(0.0480769\text{ mol}\right)\left(58.12\text{ g/mol}\right)\\ =2.7942\text{ g}\\ \approx 2.79\text{g}\end{array}$

The excess mass of ${\text{C}}_4{\text{H}}_{10}$ is calculated as follows:

  ${\text{Excess mass of C}}_4{\text{H}}_{10}=\text{Initial mass}-\text{reacted mass}$        (3)

Substitute $4.65\text{ g}$ for initial mass and $2.79\text{g}$ for reacted mass in the equation (3).

  $\begin{array}{c}\text{Excess mass of Al}=4.65\text{ g}-2.79\text{g}\\ =1.86\text{g}\text{.}\end{array}$

Conclusion

The limiting reagent controls the amount of the other reactants and the product.