Question

Recreational vehicles use propane for cooking and heating. The combustion of propane and oxygen produces carbon dioxide and water. What mass of water is produced from the combustion of \(10.0 \mathrm{~g}\) of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\) ?

Short answer

The combustion of 10.0 g of propane produces 16.36 g of water.

Step-by-step solution

Write the Balanced Chemical Equation

The balanced chemical equation for the combustion of propane is: \[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \] This shows that 1 mole of propane produces 4 moles of water.

Calculate Molar Mass of Propane

To find the number of moles of propane, first calculate its molar mass. The molecular formula is \( \text{C}_3\text{H}_8 \):\[ 3(12.01) + 8(1.01) = 44.09 \text{ g/mol} \] Thus, the molar mass of propane is \(44.09\) g/mol.

Determine Moles of Propane

Using the molar mass, calculate the number of moles of propane: \[ \text{moles of } \text{C}_3\text{H}_8 = \frac{10.0 \text{ g}}{44.09 \text{ g/mol}} = 0.227 \text{ mol} \] Thus, \(0.227\) moles of propane are present.

Calculate Moles of Water Produced

From the balanced equation, 1 mole of propane yields 4 moles of water. Thus, \(0.227\) moles of propane produce: \[ 0.227 \times 4 = 0.908 \text{ moles of } \text{H}_2\text{O} \]

Convert Moles of Water to Mass

Find the mass of \(0.908\) moles of water using its molar mass (18.02 g/mol): \[ \text{mass of } \text{H}_2\text{O} = 0.908 \text{ mol} \times 18.02 \text{ g/mol} = 16.36 \text{ g} \] Therefore, the mass of water produced is \(16.36\) g.

Key concepts

Balanced Chemical Equation

To grasp chemical reactions fully, it is crucial to start with a balanced chemical equation. This equation displays the reactants and products, ensuring that matter is conserved.
In the combustion of propane, we start with propane (\( \text{C}_3\text{H}_8 \)) and oxygen (\( \text{O}_2 \)) as the reactants, which create carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)) as products.

Consider the following steps when balancing chemical equations:

• Identify each component involved in the reaction.
• Write down their molecular formulas.
• Adjust coefficients to have the same number of each atom on both sides of the equation.

In our example, the balanced equation is: \[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \]This indicates that 1 mole of \( \text{C}_3\text{H}_8 \) reacts with 5 moles of \( \text{O}_2 \) to produce 3 moles of \( \text{CO}_2 \) and 4 moles of \( \text{H}_2\text{O} \). Understanding balanced equations helps in determining how much product forms from given reactants.

Molar Mass Calculation

The concept of molar mass allows us to convert between mass and moles, which are crucial for practical chemistry calculations.
The molar mass of a substance is the mass of one mole of its entities, usually expressed in g/mol.

When calculating the molar mass of a compound, break it down as follows:

• List each element in the compound.
• Multiply the atomic mass of each element by the number of atoms present in the compound.
• Add up these values to get the total molar mass.

For propane (\( \text{C}_3\text{H}_8 \)), we calculate it as follows:\[ \text{Molar mass of } \text{C}_3\text{H}_8 = 3(12.01) + 8(1.01) = 44.09 \text{ g/mol} \]This computation indicates that one mole of propane weighs 44.09 grams. With this method, you can determine how much of the substance you have in terms of moles, providing a foundation for further stoichiometric calculations.

Mole-to-Mole Conversion

Mole-to-mole conversion is an essential part of stoichiometry, enabling us to predict the amount of products formed in a chemical reaction based on the quantities of reactants.
In the combustion of propane, a balanced equation tells us the ratio between reactants and products. This ratio is crucial for molar conversions.

To perform a mole-to-mole conversion, follow these steps:

• Identify the given amount in moles from the chemical equation.
• Use the coefficients from the balanced equation to set up a conversion factor.
• Multiply the known moles by this conversion factor to find the desired moles.

For instance, from our balanced equation, 1 mole of \( \text{C}_3\text{H}_8 \) produces 4 moles of \( \text{H}_2\text{O} \). If we start with 0.227 moles of propane:\[ 0.227 \times 4 = 0.908 \text{ moles of } \text{H}_2\text{O} \]
This mole-to-mole relationship assists us in determining that 0.227 moles of propane result in 0.908 moles of water. Such conversions are fundamental to predicting and interpreting chemical reaction outcomes.