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Ch.3 - Chemical Reactions and Reaction Stoichiometry
Brown - Chemistry: The Central Science 15th Edition
Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 68b

The complete combustion of octane, C8H18, a component of gasoline, proceeds as follows: 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g) (b) How many grams of O2 are needed to burn 10.0 g of C8H18?

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Identify the balanced chemical equation for the combustion of octane: \(2 \text{C}_8\text{H}_{18} + 25 \text{O}_2 \rightarrow 16 \text{CO}_2 + 18 \text{H}_2\text{O}\).
Calculate the molar mass of octane (\(\text{C}_8\text{H}_{18}\)) by adding the atomic masses of carbon and hydrogen: \(8 \times 12.01 + 18 \times 1.01\).
Determine the number of moles of octane in 10.0 g by dividing the mass by the molar mass of octane.
Use the stoichiometry of the balanced equation to find the moles of \(\text{O}_2\) needed. According to the equation, 2 moles of \(\text{C}_8\text{H}_{18}\) require 25 moles of \(\text{O}_2\).
Convert the moles of \(\text{O}_2\) to grams by multiplying by the molar mass of \(\text{O}_2\) (32.00 g/mol).

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Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows chemists to calculate how much of each substance is needed or produced in a reaction based on balanced chemical equations. In this case, stoichiometry will help determine the amount of O2 required to combust a specific mass of octane.
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Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is essential for converting between the mass of a substance and the number of moles, which is necessary for stoichiometric calculations. For octane (C8H18), knowing its molar mass allows us to find out how many moles correspond to the given mass of 10.0 g.
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Balanced Chemical Equation

A balanced chemical equation represents a chemical reaction with equal numbers of each type of atom on both sides of the equation. It provides the mole ratios of reactants and products, which are crucial for stoichiometric calculations. In the combustion of octane, the balanced equation indicates the ratio of octane to oxygen, allowing us to calculate the required amount of O2 for complete combustion.
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Balancing Chemical Equations