Propane, \(C_{3} \mathrm{H}_{8}\), is a hydrocarbon that is commonly used as a fuel. (a) Write a balanced equation for the complete combustion of propane gas. (b) Calculate the volume of air at \(25^{\circ} \mathrm{C}\) and 1.00 atmosphere that is needed to completely combust 25.0 grams of propane. Assume that air is 21.0 percent \(\mathrm{O}_{2}\) by volume. (Hint: We will see how to do this calculation in a later chapter on gases - for now use the information that \(1.00 \mathrm{L}\) of air at \(25^{\circ} \mathrm{C}\) and 1.00 atm contains \(0.275 \mathrm{g}\) of \(\mathrm{O}_{2}\) per liter.) (c) The heat of combustion of propane is \(-2,219.2 \mathrm{kJ} / \mathrm{mol}\). Calculate the heat of formation, \(\Delta H_{\mathrm{f}}^{\circ}\) of propane given that \(\Delta H_{\mathrm{f}}^{\circ} \quad\) of \(\mathrm{H}_{2} \mathrm{O}(l)=-285.8 \mathrm{kJ} / \mathrm{mol}\) and \(\Delta H_{\mathrm{f}}^{\circ} \quad\) of \(\mathrm{CO}_{2}(g)=-393.5 \mathrm{kJ} / \mathrm{mol}\) (d) Assuming that all of the heat released in burning 25.0 grams of propane is transferred to 4.00 kilograms of water, calculate the increase in temperature of the water.

Step by step solution

01

Balanced Equation for Combustion of Propane

Write the balanced chemical equation for the complete combustion of propane (C_{3}H_{8}). The general form of a combustion reaction is fuel + oxygen -> carbon dioxide + water. For propane, the balanced equation is: C_{3}H_{8} + 5O_{2} -> 3CO_{2} + 4H_{2}O.

02

Calculate Moles of Propane

Find the moles of propane (C3H8) by using its molar mass (44.10 g/mol). Moles = mass (g) / molar mass (g/mol). Use the given mass of propane (25.0 g) to calculate the number of moles.

03

Determine Moles of Oxygen

Using the balanced equation, find the ratio of moles of O2 required per mole of C3H8. Then multiply this ratio by the number of moles of C3H8 calculated in Step 2 to get the moles of O2 needed.

04

Calculate Volume of Oxygen

Using the molar mass of O2 (32.00 g/mol), convert the moles of O2 to grams. Knowing that 1.00 L of air at the given conditions contains 0.275 g of O2, find the volume of O2 required for the reaction.

05

Calculate Volume of Air Required

Since the air is 21.0 percent O2 by volume, you can find the volume of air required by dividing the volume of O2 needed by the percentage of O2 in air (0.21).

06

Calculate Heat of Formation of Propane

Using Hess's Law, calculate the heat of formation for propane by using the equation for the heat of combustion and the known heats of formation for CO2 and H2O.

07

Energy Transfer to Water

Using the heat of combustion of propane and the mass of propane from part (d), calculate the total heat released. Then apply the formula q = mcΔT, where q is the heat absorbed or released, m is the mass of the water, c is the specific heat capacity of water (4.184 J/g°C), and ΔT is the change in temperature of the water.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balanced Chemical Equation

Understanding chemical reactions begins with a balanced chemical equation. The equation represents the conservation of mass principle, where the number of atoms for each element must be the same on both sides of the arrow. In the combustion of propane, we express the propane molecule, \(C_{3}H_{8}\), reacting with oxygen, \(O_{2}\), to form carbon dioxide, \(CO_{2}\), and water, \(H_{2}O\). To balance this equation for propane combustion, the coefficients indicate that one mole of propane reacts with five moles of diatomic oxygen to produce three moles of carbon dioxide and four moles of water vapor, as shown below:
\[C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(g)\]
Chemical equations must always be balanced to adhere to the law of conservation of mass. The balancing act teaches students the foundational skill of stoichiometry, helping them to quantify the relationship between reactants and products in chemical reactions.

Stoichiometry

Stoichiometry, perhaps one of the most essential concepts in chemistry, involves the calculation of the quantities of reactants and products in a chemical reaction. In our exercise, stoichiometry lets us calculate the volume of air required to burn a certain mass of propane. We start by converting the mass of propane to moles, since chemical reactions are based on moles rather than mass:
\[Moles_{propane} = \frac{Mass_{propane}}{Molar\ mass_{propane}}\]
Once we have the moles of propane, we use the balanced chemical equation to find the proportionate moles of oxygen needed, and subsequently, the volume of that oxygen. All these steps are foundational applications of stoichiometry, emphasizing its crucial role in making precise quantitative predictions in chemical reactions.

Hess's Law

Hess's Law is vital for understanding the enthalpy change in chemical reactions. It states that the total enthalpy change for a reaction is the same, regardless of the number of steps the reaction takes. This allows us to calculate the heat of formation, \(\Delta H_{\text{f}}^{\circ}\), for substances like propane. In the given exercise, using Hess's Law, we connect the enthalpies of combustion and the known heat of formation values for water and carbon dioxide to find propane's heat of formation. The beauty of Hess's Law is that it aids us in determining the change in energy without requiring the direct measurement of all reactants and products, saving time and simplifying calculations.

Energy Transfer in Reactions

Energy transfer is a key concept when examining reactions like the combustion of propane. The reaction releases energy, typically in the form of heat, and this energy can be transferred to the surroundings, such as water, raising its temperature. The change in temperature of the water is dependent on the mass of water, the specific heat capacity of water, and the heat released during the combustion process. We represent this as \(q = mc\Delta T\), where \(q\) is the energy transfer in joules, \(m\) is the water's mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. Comprehending energy transfer allows us to calculate how much energy is released or absorbed during a reaction, which is significant in both theoretical calculations and practical applications, like heating systems or engines.