/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 The complete combustion of octan... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 64

The complete combustion of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), the main component of gasoline, proceeds as follows: $$ 2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \longrightarrow 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) How many moles of \(\mathrm{O}_{2}\) are needed to burn \(1.25 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18} ?\) (b) How many grams of \(\mathrm{O}_{2}\) are needed to burn \(10.0 \mathrm{~g}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\) ? (c) Octane has a density of \(0.692 \mathrm{~g} / \mathrm{mL}\) at \(20^{\circ} \mathrm{C}\). How many grams of \(\mathrm{O}_{2}\) are required to burn \(1.00\) gal of \(\mathrm{C}_{8} \mathrm{H}_{18}\) ?

Short Answer

Expert verified
(a) \(15.625\,\text{mol}\) of \(\mathrm{O}_2\) are needed to burn \(1.25\,\text{mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\). (b) \(35.00\,\text{g}\) of \(\mathrm{O}_2\) are needed to burn \(10.0\,\text{g}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\). (c) \(9185.92\,\text{g}\) of \(\mathrm{O}_2\) are required to burn \(1.00\) gal of \(\mathrm{C}_{8} \mathrm{H}_{18}\).

Step by step solution

01

Write down the given information

We are given that 1. Moles of \(\mathrm{C}_8\mathrm{H}_{18}\) = \(1.25\,\text{mol}\).
02

Identify the stoichiometric ratio between \(\mathrm{C}_8\mathrm{H}_{18}\) and \(\mathrm{O}_2\)

From the balanced equation, we can see that the stoichiometric ratio is: $$ 2 \mathrm{C}_8\mathrm{H}_{18}: 25\mathrm{O}_2 $$
03

Calculate the moles of \(\mathrm{O}_2\) required

Since we have \(1.25\,\text{mol}\) of \(\mathrm{C}_8\mathrm{H}_{18}\), we can calculate the moles of \(\mathrm{O}_2\) required using the stoichiometric ratio: $$ \text{Moles of }\mathrm{O}_2 = \frac{25}{2}\times 1.25\,\text{mol} = 15.625\,\text{mol} $$ #(b) Grams of Oxygen Required#
04

Calculate moles of \(\mathrm{C}_8\mathrm{H}_{18}\) from given mass

To convert grams of \(\mathrm{C}_8\mathrm{H}_{18}\) to moles, use the molar mass of \(\mathrm{C}_8\mathrm{H}_{18}\) (\(\approx 114.23\,\text{g/mol}\)): $$ \text{Moles of }\mathrm{C}_8\mathrm{H}_{18} = \frac{10.0\,\text{g}}{114.23\,\text{g/mol}} \approx 0.0875\,\text{mol} $$
05

Calculate moles of \(\mathrm{O}_2\) required using stoichiometric ratio

We can use the stoichiometric ratio found in part (a) to calculate the moles of \(\mathrm{O}_2\) required: $$ \text{Moles of }\mathrm{O}_2 = \frac{25}{2}\times 0.0875\,\text{mol} \approx 1.0938\,\text{mol} $$
06

Calculate grams of \(\mathrm{O}_2\) required

To convert moles of \(\mathrm{O}_2\) to grams, use the molar mass of \(\mathrm{O}_2\) (\(\approx 32.00\,\text{g/mol}\)): $$ \text{Grams of }\mathrm{O}_2 = 1.0938\,\text{mol} \times 32.00\,\text{g/mol} \approx 35.00\,\text{g} $$ #(c) Grams of Oxygen Required to Burn 1 Gallon of Octane#
07

Convert volume of \(\mathrm{C}_8\mathrm{H}_{18}\) (gallons) to mass (grams)

We are given that \(1.00\,\text{gal}\) of \(\mathrm{C}_8\mathrm{H}_{18}\) has a density of \(0.692\,\text{g/mL}\). First, we need to convert the volume to milliliters (1 gallon \(\approx 3785.41\,\text{mL}\)): $$ 3785.41\,\text{mL} \times 0.692\,\text{g/mL} \approx 2620.12\,\text{g} $$
08

Calculate moles of \(\mathrm{C}_8\mathrm{H}_{18}\) from mass

Calculate moles of \(\mathrm{C}_8\mathrm{H}_{18}\) using the molar mass (\(\approx 114.23\,\text{g/mol}\)): $$ \text{Moles of }\mathrm{C}_8\mathrm{H}_{18} = \frac{2620.12\,\text{g}}{114.23\,\text{g/mol}} \approx 22.93\,\text{mol} $$
09

Calculate moles of \(\mathrm{O}_2\) required using stoichiometric ratio

Again, using the stoichiometric ratio found in part (a) to calculate the moles of \(\mathrm{O}_2\) required: $$ \text{Moles of }\mathrm{O}_2 = \frac{25}{2}\times 22.93\,\text{mol} \approx 287.31\,\text{mol} $$
10

Calculate grams of \(\mathrm{O}_2\) required

To convert moles of \(\mathrm{O}_2\) to grams, we use the molar mass of \(\mathrm{O}_2\) (\(\approx 32.00\,\text{g/mol}\)): $$ \text{Grams of }\mathrm{O}_2 = 287.31\,\text{mol} \times 32.00\,\text{g/mol} \approx 9185.92\,\text{g} $$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Combustion Reactions
Combustion reactions are a type of chemical reaction where a substance combines with oxygen to release energy in the form of heat and light. This process is highly exothermic, meaning it releases a significant amount of energy.
In many combustion reactions, such as the combustion of hydrocarbons like octane, the products are typically carbon dioxide (CO鈧�) and water (H鈧侽). These products illustrate the complete combustion of the organic compound.
During combustion, oxygen molecules (O鈧�) are essential as they react with the hydrocarbon. The equation for the combustion of octane is given by: \[ 2 \, ext{C}_8 ext{H}_{18}(l) + 25 \, ext{O}_2(g) \rightarrow 16 \, ext{CO}_2(g) + 18 \, ext{H}_2 ext{O}(g) \] This chemical equation shows us that 2 moles of octane react with 25 moles of oxygen to produce 16 moles of carbon dioxide and 18 moles of water. It is an important equation, as it helps us understand the stoichiometric relationships in combustion processes.
Molar Mass Calculations
Molar mass is a key component when working with chemical reactions, particularly in stoichiometry. It is the mass of one mole of a substance and is expressed in grams per mole (g/mol). Calculating the molar mass allows us to convert between moles and grams, which is crucial for determining the mass of reactants or products in a chemical reaction.
To find the molar mass of a compound, sum up the atomic masses of all the atoms in its formula. For octane (\( ext{C}_8 ext{H}_{18} \)), the molar mass is calculated as:
  • carbon (C) has an atomic mass of about 12.01 g/mol, and there are 8 carbon atoms, contributing 96.08 g/mol.
  • hydrogen (H) has an atomic mass of about 1.01 g/mol, and there are 18 hydrogen atoms, contributing 18.18 g/mol.
When combined, the molar mass of octane is approximately \( 114.23 \, ext{g/mol} \).
Molar mass calculations are essential when solving problems involving the conversion of mass to moles, as illustrated in part (b) of our exercise, where we convert a given mass of octane to moles before using the stoichiometric ratio.
Stoichiometric Ratios
Stoichiometric ratios are the proportional relationships derived from the coefficients of a balanced chemical equation. These ratios are vital for understanding how much of each reactant is required or how much of a product will be formed in a chemical reaction.
In the provided exercise, the combustion of octane uses the stoichiometric ratio from the balanced equation: \[ 2 \, ext{C}_8 ext{H}_{18} : 25 \, ext{O}_2 \] This tells us that for every 2 moles of octane, we need 25 moles of oxygen for complete combustion.
To use stoichiometric ratios, always ensure the chemical equation is balanced. Once it鈥檚 balanced, these ratios help convert known quantities, like moles of one substance, into unknown quantities, like moles of another substance. As shown in parts (a) and (c) of our exercise, this method is particularly useful for quantifying reactants and products accurately. Whether converting moles to grams or vice versa, stoichiometric ratios simplify complex chemical calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Define the terms theoretical yield, actual yield, and percent yield. (b) Why is the actual yield in a reaction almost always less than the theoretical yield? (c) Can a reaction ever have \(110 \%\) actual yield?

Calculate the percentage by mass of the indicated element in the following compounds: (a) carbon in acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\), a gas used in welding; (b) hydrogen in ascorbic acid, \(\mathrm{HC}_{6} \mathrm{H}_{7} \mathrm{O}_{6}\), also known as vitamin \(\mathrm{C}\); (c) hydrogen in ammonium sulfate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\), a substance used as a nitrogen fertilizer; (d) platinum in \(\mathrm{PtCl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\), a chemotherapy agent called cisplatin; (e) oxygen in the female sex hormone estradiol, \(\mathrm{C}_{18} \mathrm{H}_{24} \mathrm{O}_{2} ;\) (f) carbon in capsaicin, \(\mathrm{C}_{18} \mathrm{H}_{27} \mathrm{NO}_{3}\), the com- pound that gives the hot taste to chili peppers.

An organic compound was found to contain only C, \(\mathrm{H}\), and \(\mathrm{Cl}\). When a \(1.50-\mathrm{g}\) sample of the compound was completely combusted in air, \(3.52 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) was formed. In a separate experiment the chlorine in a 1.00-g sample of the compound was converted to \(1.27 \mathrm{~g}\) of \(\mathrm{AgCl}\). Determine the empirical formula of the compound.

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) together with other substances. Reaction of the ore with CO produces iron metal: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ (a) Balance this equation. (b) Calculate the number of grams of CO that can react with \(0.150 \mathrm{~kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). (c) Calculate the number of grams of Fe and the number of grams of \(\mathrm{CO}_{2}\) formed when \(0.150 \mathrm{~kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

Aluminum hydroxide reacts with sulfuric acid as follows: $$ 2 \mathrm{Al}(\mathrm{OH})_{3}(s)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ Which reagent is the limiting reactant when \(0.500 \mathrm{~mol}\) \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(0.500 \mathrm{~mol} \mathrm{H}_{2} \mathrm{SO}_{4}\) are allowed to react? How many moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.