Question

Bottled propane is used in areas away from natural gas pipelines for cooking and heating, and is also the source of heat in most gas barbecue grills. Propane burns in oxygen according to the following balanced chemical equation: $$\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)$$ Calculate the mass in grams of water vapor produced if 3.11 mol of propane is burned.

Short answer

The mass of water vapor produced when 3.11 moles of propane is burned is approximately 223.97 grams.

Step-by-step solution

Find the number of moles of water vapor produced

Using the balanced chemical equation: C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g) We can see that 1 mole of propane produces 4 moles of water vapor. Therefore, for 3.11 moles of propane burned, we can use the following proportion: \( \frac{4 \text{ moles of water vapor}}{1 \text{ mole of propane}} = \frac{x \text{ moles of water vapor}}{3.11 \text{ moles of propane}}\) Solve for x: \( x = 4 \times 3.11 \)

Calculate the number of moles of water vapor

Now, calculate x (moles of water vapor): \(x = 4 \times 3.11 = 12.44 \; \mathrm{moles} \; \) So, 12.44 moles of water vapor are produced.

Find the mass of water vapor

To find the mass of water vapor produced, we need to know the molar mass of water (H₂O). The molar mass of H₂O is approximately 18.015 g/mol (16.00 g/mol for O and 1.008 g/mol for H x 2). Now, use the moles of water vapor (12.44 moles) and the molar mass to calculate the mass of water vapor produced: mass of water vapor = moles of water vapor × molar mass of H₂O = 12.44 moles × 18.015 g/mol

Calculate the mass of water vapor

Now, perform the calculation for the mass of water vapor: mass of water vapor = 12.44 moles × 18.015 g/mol ≈ 223.97 g Thus, approximately 223.97 grams of water vapor will be produced if 3.11 moles of propane is burned.

Key concepts

Propane Combustion

Propane combustion is a chemical reaction where propane (\(\text{C}_3\text{H}_8\)) reacts with oxygen to produce carbon dioxide and water vapor. This reaction is crucial for burning fuel in various heating applications, from outdoor grills to home heating systems in areas without natural gas pipelines.

When propane combusts, it releases heat energy, which is why it's a popular source of fuel.

• Propane needs a sufficient supply of oxygen to burn cleanly, producing \(\text{CO}_2\) and \(\text{H}_2\text{O}\).
• If oxygen is inadequate, it can produce carbon monoxide (\(\text{CO}\)), which is dangerous.

Understanding this process helps ensure safety and efficiency in using propane-fueled appliances.

Balanced Chemical Equation

A balanced chemical equation ensures that the number of atoms for each element is the same on both sides of the reaction. This is based on the law of conservation of mass, stating that mass cannot be created or destroyed.

For the combustion of propane:\[\text{C}_3\text{H}_8(g) + 5\,\text{O}_2(g) \rightarrow 3\,\text{CO}_2(g) + 4\,\text{H}_2\text{O}(g)\]This equation shows that:

• 1 molecule of propane reacts with 5 molecules of oxygen.
• The reaction forms 3 molecules of carbon dioxide and 4 molecules of water vapor.

Balancing chemical equations requires counting and matching atoms on both sides, ensuring the reaction is valid and accurately represents reality.

Molar Mass Calculation

Molar mass calculation helps convert between moles and grams, the common unit in the laboratory. It is crucial in stoichiometry, which involves the quantitative relationship between reactants and products in a chemical reaction.

Taking water (\(\text{H}_2\text{O}\)) as an example:

• Hydrogen (\(\text{H}\)) has a molar mass of approximately 1.008 g/mol.
• Oxygen (\(\text{O}\)) has a molar mass of about 16.00 g/mol.
• Therefore, water's molar mass is \(2 \times 1.008 + 16.00 = 18.015 \, \text{g/mol}\).

This calculation allows us to find the mass of substances produced or consumed in a reaction, like determining that 12.44 moles of water vapor equate to approximately 223.97 grams when propane burns.