A liquid fuel mixture contains 30.35% hexane (C 6 H 14 ), 15.85% heptane (C 7 H 16 ), and the rest octane (C 8 H 18 ). What maximum mass of carbon dioxide is produced by the complete combustion of 10.0 kg of this fuel mixture?

Welcome back everyone. A liquid fuel mixture contains 31.25% pens 17.43% hexane and the rest no name. What is the maximum mass of carbon dioxide produced by the complete combustion of 10.0 kg of this fuel mixture? And we are given for answer choices. A 22.8 B 28.1 C 30.7 and D 36.1 all given in kilograms. So essentially, first of all, we have to understand that a complete combustion means that we are burning every of our hydrocarbons in oxygen to produce carbon dioxide and water. So let's start with pens. Let's analyze the reaction equation C five H 12. That's Beane. It reacts with oxygen to produce carbon dioxide and a water vapor. If we balance the equation, we have to add five for CO2. Now, since we have 12 hydrogens, we will add six for water, the total number of oxygens would be six plus 10. That's 16. So we want to add eight in front of 02. Now, what about hexane C 614? We once again add oxygen, we produce co two and water. If we balance it, we immediately add six for CO2 7 4h to the total number of oxygens would be 12 plus seven which is 19. So we have 19 divided by two oxygen. Ok? Now no name C nine age 20 plus oxygen. We release CO two and H2O. So now we add nine for CO2 10 for HTO. The total number of oxygens would be 10 plus 18. That's 28. So we need 14 molecules of +02. Well done. Now, what we're going to do is just calculate the masses of CO2 released by each reaction. OK. To do that, we first will need to get the mass of Pensa. We understand that we have 31.25% of pens and the total mass of the sample is 10 kg. So we can say that mass one of CO2, right, we're going to calculate the mass of CO2 released by the first reaction. We can label it as one followed by reaction number two and three. So to get the first mass of CO two, we're going to take 10 kg is the total mass of the sample. We're going to multiply that by 31.25% which in a decimal form is 0.3125. This gives us the kilograms of pens in the given sample. We're going to convert that into grams knowing that there are 1000 g in 1 kg. OK? And now from here, we're going to calculate moles, right? So first of all, we need to include one mole on top and on the bottom, we're going to include the molar mass of insane where Ben saying the molar mass would be 72.15 grams per mole. Now, we're going to use the geometry to get the number of moles of CO2, right? So we know that we have five moles of CO two for every one mole of C five H 12. Now we have the number of moles of CO2. We want to get the mass of CO two, right. So we're going to multiply by the molar mass of CO2, 44.01 g in one mole. And from here, we simply want to convert the mass of CO2 into kilograms. We know that there's 1 kg in a 1000 g well done. So we can get the mass of CO two in this stuff. We get nine points 5 3 kg. Let's continue. The mass of CO two coming from the second reaction would be 7.0 kg multiplied by the percentage of hexane. 011743. We are going to convert that into grams. Now, into molds of hexane, the molar mass of hexane would be 86.17 grams per mole. Now, according to ST geometry, there are six moles of CO2 for every mole of hexane. We're multiplying by the molar mass of carbon dioxide and now converting the answer into kilograms. Now let's see what we got in this step, the mass of CO2 would be five point 34 kilograms. And finally, we want to get the mass of CO two coming from the third reaction. So once again, we take 10 kg, we know that the rest of the sample is no name. So we're going to take one minus 0.3125 minus 0.1743. And this gives us a fraction of nona right, we have 100% of the sample and we have to subtract the percentages of Beane and hexane. Now we convert that into grams. Now into moles. Let's use the molar mass of non which is 128.2 g per mole. Now we know that the ratio according to ST geometry is 9 to 1. So we take multiplied by nine moles for every mole of no name. We have the number of moles of carbon dioxide. So we are multiplying by the molar mass of carbon dioxide to get grams of carbon dioxide. And from here we are converting grams into kilograms. OK? And now if we calculate the answer for this step, we get 15 0.85 kilograms. Now if we want to get the total mass, we have to add the mass obtained in step one, step two and step three. When we add those three numbers specifically 9.53 kg 5.34 kg and 15.85 kg. We get 30.7 kg, meaning the correct answer to this problem would be option C 30.7 kg of CO2 would be produced. Thank you for watching.

Ch.4 - Chemical Reactions and Chemical Quantities

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Ch.4 - Chemical Reactions and Chemical Quantities

Problem 67

Chapter 4, Problem 67

A liquid fuel mixture contains 30.35% hexane (C₆H₁₄), 15.85% heptane (C₇H₁₆), and the rest octane (C₈H₁₈). What maximum mass of carbon dioxide is produced by the complete combustion of 10.0 kg of this fuel mixture?

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Determine the percentage of octane (C_8H_18) in the mixture by subtracting the percentages of hexane and heptane from 100%.

Calculate the mass of each component (hexane, heptane, and octane) in the 10.0 kg fuel mixture using their respective percentages.

Write the balanced chemical equations for the complete combustion of hexane, heptane, and octane. For example, the combustion of hexane is: C_6H_14 + \frac{19}{2}O_2 \rightarrow 6CO_2 + 7H_2O.

Use stoichiometry to calculate the moles of CO_2 produced from the complete combustion of each component. Use the molar mass of each hydrocarbon to convert mass to moles, and then use the balanced equations to find moles of CO_2.

Convert the total moles of CO_2 produced from all components to mass using the molar mass of CO_2 (44.01 g/mol) to find the maximum mass of CO_2 produced.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Combustion Reactions

Combustion reactions involve the reaction of a fuel with oxygen to produce carbon dioxide and water. In the case of hydrocarbons like hexane, heptane, and octane, the general reaction can be represented as C_nH_m + O_2 → CO_2 + H_2O. Understanding the stoichiometry of these reactions is essential for calculating the products formed from a given amount of fuel.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It allows us to determine how much of each substance is involved in the reaction based on balanced chemical equations. For this question, stoichiometry will be used to calculate the amount of carbon dioxide produced from the combustion of the fuel mixture.

Molar Mass and Mass Percentages

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole. In this question, the mass percentages of hexane, heptane, and octane in the fuel mixture are used to determine the total mass of each component in the 10.0 kg mixture. This information is crucial for calculating the total amount of carbon dioxide produced during combustion.