The complete combustion of octane, C8H18, a component of gasoline, proceeds as follows: 2 C 8 H 18 (l) + 25 O 2 (g) → 16 CO 2 (g) + 18 H 2 O(g) (b) How many grams of O 2 are needed to burn 10.0 g of C 8 H 18 ?

Alright folks, so we have a balanced reaction below here that shows the combustion of propane. Right? So we have propane combusting with oxygen producing C. 02 and water. So if we're told that we have 132 point grams of C three H eight of propane, that's completely combusted. We need to calculate the mass in grams of oxygen needed for this process. So we want Massive oxygen given the mass of propane. Okay, so let's go ahead and start. So we already told that this is a balanced reaction. So we don't need to check. But you can go ahead and check anyways, it is balanced and we are ready to start. So 132.1 grams of propane. That's what we're going to start with. Because that's the only thing that we have given to us that we know. And we're gonna go ahead and convert this into moles so that we can do a multiple comparison between propane and oxygen to find uh the mass of 02. Alright, so in one mole of propane, how many grams do we have? That is just the molar mass of propane. Let's go ahead and find that on the side. So we have carbon and hydrogen For the carbon, we have three of them. Right? And each of them weigh 12. So that gives us 36 Gramps. And then for the hydrogen is we have eight of them. They weigh 1.01 each. So that is 8.08 g. And then in total, this will be 44.08 grams Permal. So that's the molar mass of propane. So we're going to put that on the bottom. So 44.08 grams of C three H eight In one mole of it. Okay, so g cancel out. And now we have moles. Now we can do a multiple comparison, right? So we can see that there is only one mole of propane. We're going to put that on the bottom so that it can cancel out. So for every one mole of propane we're using five moles of 02. So moles of O to go on the top because that's what we want. Right? So now that we have moles of 02, we can very easily convert that into grams right into mass and grams. So again, we just need the molar mass of 02. So in one mole of 02 and moles of O. To go on the bottom. So that we can cancel that out. How many grams are there? Well as oxygen, it's 16 g. Right? There's two of them. 16 times two is 32. So there are 32 g of 02 in one mole of 02. So moles cancel out and now we have grams of 02. And that will be our answer. So let's go ahead and divide and multiply everything correctly here. And it should give you 479.5 g of co two. Alright, so that's gonna be our answer for the mass and grams of CO. Two needed for this combustion reaction. Thank you so much for watching.

Ch.3 - Chemical Reactions and Reaction Stoichiometry

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Brown 14th Edition

Ch.3 - Chemical Reactions and Reaction Stoichiometry

Problem 68b

Chapter 3, Problem 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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Key Concepts

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

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.

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.

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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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) (a) How many moles of O₂ are needed to burn 1.50 mol of C₈H₁₈?

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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?

Verified video answer for a similar problem:

Key Concepts

Stoichiometry

Molar Mass

Balanced Chemical Equation

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?