Question

How many grams of ${\mathrm{CO}}_2$ are produced in the complete combustion of $406\mathrm{g}$ of a bottled gas that consists of $72.7\mathrm{\%}$ propane $\left({\mathrm{C}}_3{\mathrm{H}}_8\right)$ and $27.3\mathrm{\%}$ butane $\left({\mathrm{C}}_4{\mathrm{H}}_{10}\right)$ by mass?

Short answer

The total amount of ${\mathrm{CO}}_2$ produced from the combustion reaction of the bottled gas is 1184.33 g.

Step-by-step solution

Determine the amount of propane and butane

First, calculate the amount of propane (${\mathrm{C}}_3{\mathrm{H}}_8$) and butane (${\mathrm{C}}_4{\mathrm{H}}_{10}$) in grams in the bottled gas using their given percentages respectively. This will give $406g\times 0.727$ for propane and $406g\times 0.273$ for butane.

Write down the balanced combustion reaction

The balanced chemical equations for the combustion of propane and butane are ${\mathrm{C}}_3{\mathrm{H}}_8+5{\mathrm{O}}_2\to 3{\mathrm{CO}}_2+4{\mathrm{H}}_2\mathrm{O}$ and ${\mathrm{C}}_4{\mathrm{H}}_{10}+6.5{\mathrm{O}}_2\to 4{\mathrm{CO}}_2+5{\mathrm{H}}_2\mathrm{O}$ respectively. Notice that in both equations, number of carbon dioxide molecules produced is equal to the number of carbon atoms in the reactant hydrocarbon molecule.

Calculate the amount of CO2 produced

For propane, which has 3 carbon atoms in a molecule, this means we will get 3 moles of ${\mathrm{CO}}_2$ from 1 mole of ${\mathrm{C}}_3{\mathrm{H}}_8$, and for butane with 4 carbon atoms in a molecule, we will get 4 moles of ${\mathrm{CO}}_2$ from 1 mole of ${\mathrm{C}}_4{\mathrm{H}}_{10}$. To get moles from grams, use the molecular weights (44.1 g/mol for ${\mathrm{CO}}_2$, 44.1 g/mol for ${\mathrm{C}}_3{\mathrm{H}}_8$, and 58.12 g/mol for ${\mathrm{C}}_4{\mathrm{H}}_{10}$). Multiply the moles of each hydrocarbon by the corresponding conversion factor to ${\mathrm{CO}}_2$ moles. Then, convert back to grams using the molecular weight of ${\mathrm{CO}}_2$.

Add up the two amounts

Finally add the amounts for ${\mathrm{CO}}_2$ produced from both propane and butane to get the total amount of ${\mathrm{CO}}_2$ produced.

Key concepts

Combustion Reaction

When a substance burns, it usually reacts with oxygen in the air to produce heat and light, a process known as combustion. In a combustion reaction, the reactants are typically hydrocarbons and oxygen. The products are usually carbon dioxide and water. For propane (${\text{C}}_3{\text{H}}_8$) and butane (${\text{C}}_4{\text{H}}_{10}$), their combustion reactions can be written as balanced chemical equations:

• Propane: ${\text{C}}_3{\text{H}}_8+5{\text{O}}_2\to 3{\text{CO}}_2+4{\text{H}}_2\text{O}$
• Butane: ${\text{C}}_4{\text{H}}_{10}+6.5{\text{O}}_2\to 4{\text{CO}}_2+5{\text{H}}_2\text{O}$

During combustion of these hydrocarbons, the carbon atoms from the hydrocarbon react with oxygen molecules. This produces carbon dioxide.To ensure complete reactions, enough oxygen must be present. Make sure you understand how to balance chemical reactions to check the stoichiometry of the reactants and products.

Moles and Molar Mass

In chemistry, amounts of substances are often measured in moles. A mole is a basic counting unit, similar to a dozen, but instead of 12 it represents Avogadro's number, approximately $6.022\times {10}^{23}$entities. One mole of any substance contains this same number of molecules or atoms.

The molar mass is the weight of one mole of a substance and is usually given in grams per mole (g/mol). It is calculated by adding together the atomic masses of all the atoms in a molecule.

For example, the molar mass of carbon dioxide (${\text{CO}}_2$) is calculated as follows:

• Carbon: $12.01\text{ g/mol}$
• Oxygen: $16.00\text{ g/mol}$ (since there are two oxygen atoms: $2\times 16.00=32.00\text{ g/mol}$)
• Total molar mass of ${\text{CO}}_2=44.01\text{ g/mol}$

For propane (${\text{C}}_3{\text{H}}_8$), each molecule has 3 carbon atoms and 8 hydrogen atoms. The molar mass is:

• 3 Carbon: $3\times 12.01=36.03\text{ g/mol}$
• 8 Hydrogen: $8\times 1.01=8.08\text{ g/mol}$
• Total: $44.11\text{ g/mol}$

Knowing how to find moles using mass and molar mass is crucial for solving chemistry problems.

Stoichiometry

In stoichiometry, we use balanced chemical equations to determine the relationships between reactants and products in a chemical reaction. This allows us to calculate the amount of product formed based on the amounts of reactants used.

For the combustion of propane and butane, stoichiometry involves using the mole ratios from the balanced equations. For instance, if you start with 1 mole of propane, the equation tells you that you will get 3 moles of ${\text{CO}}_2$. The mole ratio between ${\text{C}}_3{\text{H}}_8$ and ${\text{CO}}_2$ is 1:3, and for butane, the ratio is 1:4. This means:

• 1 mole of propane yields 3 moles of ${\text{CO}}_2$.
• 1 mole of butane yields 4 moles of ${\text{CO}}_2$.

To find the total grams of ${\text{CO}}_2$ produced:

• First, calculate the moles of propane and butane in the 406g mixture.
• Convert the grams of each gas to moles using their molar masses.
• Apply the stoichiometric ratios to find the moles of ${\text{CO}}_2$ produced.
• Finally, convert these moles of ${\text{CO}}_2$ back to grams using its molar mass.

Stoichiometry is an essential concept for predicting amounts of products and for ensuring that reactions are both efficient and safe in practical applications.