Problem 70 - Chemistry: Structure and Properties

Question

Many home barbeques are fueled with propane gas (C3H8). What mass of carbon dioxide (in kg) forms upon the complete combustion of 18.9 L of propane (approximate contents of one 5-gallon tank)? Assume that the density of the liquid propane in the tank is 0.621 g>mL. (Hint: Begin by writing a balanced equation for the combustion reaction.)

Step-by-Step Solution

Answer

The mass of carbon dioxide formed upon the complete combustion of 18.9 L of propane is approximately 35.14 kg.

Step 1: Write the balanced chemical equation

The combustion of propane (\text{C}_3\text{H}_8) in the presence of oxygen (\text{O}_2) produces carbon dioxide (\text{CO}_2) and water (\text{H}_2\text{O}). The balanced chemical equation is: \[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \]

Step 2: Calculate the mass of propane

First, convert the volume of propane to its mass by using the given density. \[ \text{mass of propane} = \text{volume} \times \text{density} \] \[ \text{mass of propane} = 18.9\text{ L} \times 1000\text{ mL/L} \times 0.621\text{ g/mL} \] \[ \text{mass of propane} = 11739\text{ g} \]

Step 3: Convert the mass of propane to moles

Use the molar mass of propane (C_3H_8, which is 44.10 g/mol) to convert the mass to moles. \[ \text{moles of propane} = \frac{\text{mass of propane}}{\text{molar mass of propane}} \] \[ \text{moles of propane} = \frac{11739\text{ g}}{44.10\text{ g/mol}} \] \[ \text{moles of propane} = 266.19\text{ mol} \]

Step 4: Use stoichiometry to find moles of CO2 produced

From the balanced equation, 1 mole of C_3H_8 produces 3 moles of CO_2. \[ \text{moles of CO}_2 = \text{moles of propane} \times 3 \] \[ \text{moles of CO}_2 = 266.19 \times 3 \] \[ \text{moles of CO}_2 = 798.57 \]

Step 5: Convert moles of CO2 to mass

Use the molar mass of CO_2 (44.01 g/mol) to convert the moles of CO_2 to mass in grams and then convert grams to kilograms. \[ \text{mass of CO}_2 = \text{moles of CO}_2 \times \text{molar mass of CO}_2 \] \[ \text{mass of CO}_2 = 798.57\text{ mol} \times 44.01\text{ g/mol} \] \[ \text{mass of CO}_2 = 35139.58\text{ g} \] \[ \text{mass of CO}_2 \text{ in kg }= \frac{35139.58\text{ g}}{1000\text{ g/kg}} \] \[ \text{mass of CO}_2 \text{ in kg }= 35.14\text{ kg} \]

Key Concepts

Stoichiometry

Stoichiometry is a section of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is essentially the math behind chemistry, allowing us to predict how much of each substance is involved in a reaction. This concept is crucial for the combustion of propane, where we use the balanced chemical equation to determine the amount of carbon dioxide produced.

When solving a stoichiometry problem, such as the combustion of propane in the exercise, it begins with the balanced chemical equation. In this case, one mole of propane reacts with five moles of oxygen to produce three moles of carbon dioxide and four moles of water. Using the coefficients from the balanced equation, we handle the ratios of substances quantitatively. For the propane combustion, three times the number of moles of propane will be the moles of carbon dioxide produced, simply because of the stoichiometric coefficient 3 in front of the CO₂ in the equation.

Chemical Reactions

A chemical reaction is a process that leads to the transformation of one set of chemical substances to another. During a chemical reaction, the substances you start with—reactants—are used up to create new substances—products. An understanding of chemical reactions is fundamental to solving problems relating to the combustion of propane.

For the exercise given, we specifically look at a combustion reaction, which is a type of exothermic reaction where a substance combines with oxygen, releasing energy mostly in the form of heat and light. Propane combustion involves an initial momentum where propane molecules and oxygen molecules collide to form carbon dioxide and water as products. These products are entirely different at a molecular level from the reactants, depicting a profound change, which is the essence of any chemical reaction.

Molar Mass

The molar mass of a substance is the mass of one mole of that substance and is usually expressed in grams per mole (g/mol). It is an essential concept for converting between the mass of a substance and the amount in moles, which is key for stoichiometry calculations. Knowing the molar mass allows us to convert grams to moles, which can then be used in stoichiometric calculations to find coefficients in a balanced equation.

In the combustion problem, we use the molar mass of propane (44.10 g/mol) to convert the given mass of propane into moles. This step is pivotal as it transitions the quantity from a tangible mass (which we can measure or be given as in the problem) to the mole unit, which then directly relates to the chemical equation and stoichiometric ratios. The molar mass of carbon dioxide (44.01 g/mol) is similarly utilized to find the final mass of CO₂ produced from the moles of CO₂. Without accurate molar mass values, predictions regarding the outcome of a reaction cannot be made precisely.