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The complete combustion of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), produces \(5470 \mathrm{~kJ}\) of heat. Calculate how many grams of octane is required to produce \(20,000 \mathrm{~kJ}\) of heat.

Short Answer

Expert verified
You need 835.38 grams of octane to produce 20,000 kJ of heat.

Step by step solution

01

Write the Chemical Reaction Equation

The complete combustion of octane, \(\text{C}_8\text{H}_{18}\), with oxygen produces carbon dioxide and water. The balanced chemical equation is: \[ 2\text{C}_8\text{H}_{18} + 25\text{O}_2 \rightarrow 16\text{CO}_2 + 18\text{H}_2\text{O} \]
02

Determine Heat Produced Per Mole of Octane

We know \(5470 \, \text{kJ}\) of heat is produced from the combustion of octane. Since the balanced equation refers to the combustion of 2 moles of octane, the heat produced per mole of octane is \( \frac{5470}{2} \, \text{kJ/mol}\), which equals \(2735 \, \text{kJ/mol}\).
03

Calculate Moles of Octane Needed

We want to produce \(20,000 \, \text{kJ}\). Thus, we calculate the moles of octane needed as follows: \[ x = \frac{20,000}{2735} \approx 7.313 \, \text{mol} \]
04

Convert Moles of Octane to Grams

The molar mass of octane (\(\text{C}_8\text{H}_{18}\)) is calculated by summing the atomic masses: \[ 8(12.01) + 18(1.01) = 114.23 \, \text{g/mol} \] To find the mass of octane needed, multiply the moles by the molar mass: \[ 7.313 \times 114.23 = 835.38 \, \text{g} \]

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Key Concepts

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

Understanding Heat Calculation in Combustion Reactions
The process of calculating heat involves understanding the energy changes that occur during a chemical reaction. In the case of a combustion reaction, energy is released as heat. For octane combustion, the reaction releases a specific amount of heat energy. From the problem, we know that burning octane generates \(5470 \, \text{kJ}\) for every two moles. To determine how much octane is required to generate \(20,000 \, \text{kJ}\) of heat, we use a simple proportion. First, calculate the heat released per mole of octane by dividing the total heat \(5470 \, \text{kJ}\) by 2, resulting in \(2735 \, \text{kJ/mol}\). This ratio tells us the heat produced per one mole. To find the moles needed to reach \(20,000 \, \text{kJ}\), divide \(20,000 \, \text{kJ}\) by the \(2735 \, \text{kJ/mol}\), resulting in approximately \(7.313 \, \text{mol}\). This amount equates to the production of the desired heat value through the combustion of octane.
The Role of Chemical Reaction Equations
A chemical reaction equation represents the reactants and products involved in a chemical reaction. For the combustion of octane, the equation is:\[ 2\text{C}_8\text{H}_{18} + 25\text{O}_2 \rightarrow 16\text{CO}_2 + 18\text{H}_2\text{O} \]This equation indicates that two moles of octane react with twenty-five moles of oxygen gas. This produces sixteen moles of carbon dioxide and eighteen moles of water. Balanced chemical equations are essential. They ensure that the number of atoms for each element is the same on both sides of the equation. This balance maintains the principle of conservation of mass. By understanding and utilizing these equations, we can accurately predict the amounts of reactants consumed and products formed. In this scenario, the balanced equation was crucial for determining how much heat is produced from a specific quantity of octane.
Calculating Molar Mass for Conversion to Grams
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole \(\text{g/mol}\). It's a fundamental concept used to convert between moles and grams, which are practical units for measuring chemical substances. To find the molar mass of octane (\(\text{C}_8\text{H}_{18}\)), we sum the atomic masses of its constituent atoms. Octane has eight carbon atoms, each with an atomic mass of approximately \(12.01 \, \text{g/mol}\), and eighteen hydrogen atoms, each weighing about \(1.01 \, \text{g/mol}\). Putting it together:
  • Carbon: \(8 \times 12.01 = 96.08 \, \text{g/mol}\)
  • Hydrogen: \(18 \times 1.01 = 18.18 \, \text{g/mol}\)
Add them to get the total molar mass of octane: \(114.23 \, \text{g/mol}\).This molar mass is pivotal in converting from moles (\(7.313 \, \text{mol}\) as calculated) to grams by multiplying the number of moles by the molar mass. Therefore, \(7.313 \, \text{mol} \times 114.23 \, \text{g/mol}\) equals approximately \(835.38 \, \text{g}\). This calculation shows how much octane is needed to produce \(20,000 \, \text{kJ}\) of heat.

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Most popular questions from this chapter

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of \(\mathrm{CO}, 0.733 \mathrm{~g}\) of \(\mathrm{CO}_{2}\), and \(0.450 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing \(3.50 \mathrm{~g}\) of sodium carbonate is mixed with one containing \(5.00 \mathrm{~g}\) of silver nitrate. How many grams of sodium carbonate, silver nitrate, silver carbonate, and sodium nitrate are present after the reaction is complete?

Determine the formula weights of each of the following compounds: (a) lead (IV) chloride; (b) copper(II) oxide; (c) iodic acid, \(\mathrm{HIO}_{3} ;(\mathbf{d})\) sodium perchlorate, \(\mathrm{NaClO}_{4}\); (e) indium nitride, (f) phosphorus pentoxide, \(\mathrm{P}_{4} \mathrm{O}_{10} ;(\mathbf{g})\) boron trichloride.

Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: (a) \(\mathrm{NaClO}_{3}(s) \longrightarrow \mathrm{NaCl}(s)+\mathrm{O}_{2}(g)\) (b) \(\mathrm{NH}_{4} \mathrm{OH}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NH}_{3}(g)\) (c) \(\mathrm{K}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{KCl}(s)\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(l)\)

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. If \(5.00 \mathrm{~g}\) of sulfuric acid and \(5.00 \mathrm{~g}\) of lead(II) acetate are mixed, calculate the number of grams of sulfuric acid, lead(II) acetate, lead(II) sulfate, and acetic acid present in the mixture after the reaction is complete.

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