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Chapter 3: Problem 115

Propane burns according to the following equation: $$ \mathrm{C}_{3} \mathrm{H}_{8}+5 \mathrm{O}_{2} \longrightarrow 3 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O} $$ (a) How many grams of \(\mathrm{O}_{2}\) are needed to react fully with \(3.45 \mathrm{~mol}\) of propane? (b) How many grams of \(\mathrm{CO}_{2}\) can form from \(0.177 \mathrm{~mol}\) of propane? (c) How many grams of water are produced by the combustion of \(4.86 \mathrm{~mol}\) of propane?

Short Answer

Expert verified
For part (a), 552.00 g of \(O_2\) are needed; for part (b), 23.37 g of \(CO_2\) can form; for part (c), 350.25 g of water are produced.

Step by step solution

01

Balance the combustion equation

First, verify that the chemical equation for the combustion of propane is balanced. It is given as \(C_3H_8 + 5O_2 \longrightarrow 3CO_2 + 4H_2O\). This equation is already balanced.
02

Convert moles of propane to moles of O2 for part (a)

Use the mole ratio from the balanced equation to find the moles of \(O_2\) needed. The ratio is \(5 mol \, O_2 / 1 mol \, C_3H_8\). Multiply the moles of propane (3.45 mol) by this ratio: \(3.45 mol \, C_3H_8 \times (5 mol \, O_2 / 1 mol \, C_3H_8) = 17.25 mol \, O_2\).
03

Convert moles of O2 to grams for part (a)

Convert moles of \(O_2\) to grams using the molar mass of \(O_2\) (32.00 g/mol): \(17.25 mol \, O_2 \times 32.00 g/mol = 552.00 g \, O_2\).
04

Convert moles of propane to moles of CO2 for part (b)

Use the mole ratio to find the moles of \(CO_2\) produced. The ratio is \(3 mol \, CO_2 / 1 mol \, C_3H_8\). Multiply the moles of propane (0.177 mol) by this ratio: \(0.177 mol \, C_3H_8 \times (3 mol \, CO_2 / 1 mol \, C_3H_8) = 0.531 mol \, CO_2\).
05

Convert moles of CO2 to grams for part (b)

Convert moles of \(CO_2\) to grams using the molar mass of \(CO_2\) (44.01 g/mol): \(0.531 mol \, CO_2 \times 44.01 g/mol = 23.37 g \, CO_2\).
06

Convert moles of propane to moles of H2O for part (c)

Use the mole ratio to find the moles of \(H_2O\) produced. The ratio is \(4 mol \, H_2O / 1 mol \, C_3H_8\). Multiply the moles of propane (4.86 mol) by this ratio: \(4.86 mol \, C_3H_8 \times (4 mol \, H_2O / 1 mol \, C_3H_8) = 19.44 mol \, H_2O\).
07

Convert moles of H2O to grams for part (c)

Convert moles of \(H_2O\) to grams using the molar mass of \(H_2O\) (18.02 g/mol): \(19.44 mol \, H_2O \times 18.02 g/mol = 350.25 g \, H_2O\).

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Key Concepts

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

Stoichiometry Calculations
Stoichiometry calculations are a fundamental aspect of chemistry that deal with the quantitative relationship between reactants and products in a chemical reaction. These calculations are based on the balanced chemical equation which provides the mole ratios between the compounds involved.

For instance, in the combustion of propane, the balanced equation \(C_3H_8 + 5O_2 \longrightarrow 3CO_2 + 4H_2O\) tells us that one mole of propane reacts with five moles of oxygen to produce three moles of carbon dioxide and four moles of water. To solve real-world problems, we often start with a certain amount of reactant鈥攊n moles, grams, or liters鈥攁nd use stoichiometry to predict the amount of product produced or reactant needed.

Molar Mass Conversion
Molar mass conversion is crucial for translating between the mass of a substance and the amount in moles, bridging the gap between the macroscopic world (grams) and the microscopic (molecular scale). The molar mass is defined as the mass of one mole of a substance and has units of grams per mole (g/mol).

Molar mass is used as a conversion factor in stoichiometry calculations to convert moles to grams or vice versa. For example, oxygen has a molar mass of 32.00 g/mol. Therefore, if the combustion of propane requires 17.25 moles of oxygen, we convert this to grams by multiplying it by the molar mass of oxygen, resulting in 552.00 grams.

Balanced Chemical Equations
A balanced chemical equation is one in which the number of atoms of each element is the same on both sides of the equation, adhering to the Law of Conservation of Mass. This balance is necessary for the accurate use of stoichiometry in chemical calculations.

The equation for the combustion of propane \(C_3H_8 + 5O_2 \longrightarrow 3CO_2 + 4H_2O\) shows a balanced reaction where the atoms of carbon, hydrogen, and oxygen are conserved through the process. To achieve this, coefficients are used to multiply the number of molecules, hence providing the correct stoichiometric ratios which are essential for stoichiometry calculations, allowing us to relate moles of propane to moles of oxygen, carbon dioxide, and water.

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Most popular questions from this chapter

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