How many grams CO2 form from the complete combustion of 1.00 L C8 H18, density 0.700 g / mL ? If

How many grams CO2 form from the complete combustion of 1.00...

Key Concepts

Percentage Yield

Percentage yield is a measure of the efficiency of a chemical reaction, comparing the actual yield obtained from an experiment to the theoretical yield predicted by stoichiometric calculations. This concept helps in understanding practical limitations in chemical reactions and quantifying the extent to which a reaction proceeds.

Combustion Reaction

A combustion reaction is a type of chemical reaction where a substance, typically a hydrocarbon, reacts with oxygen to produce carbon dioxide and water. Understanding the general nature of combustion reactions is crucial for setting up the stoichiometric relationships necessary to determine the amounts of products formed.

Density

Density is a physical property of a substance that relates its mass to its volume. In quantitative problems, density is used to convert between volume and mass, which is especially useful when a substance’s amount is given in volume and the calculations require mass.

Stoichiometry

Stoichiometry involves using a balanced chemical equation to relate the quantities of reactants and products in a chemical reaction. It allows for the calculation of the amount of product formed from a specified amount of reactant by using mole ratios derived from the coefficients in the balanced equation.

Molar Mass

Molar mass is the mass of one mole of a substance and is used to convert between the mass of a substance and the number of moles. This conversion is essential in stoichiometry as it bridges the gap between the macroscopic measurement of mass and the microscopic counting of atoms or molecules.

Transcript

00:01 Okay, so we, this is a percent yield problem, and we are given the combustion of octane, once again, and we're told that we observe the formation of 1 .90 times 10 to the 3rd grams of co2.

00:17 We have to figure out the percent yield of this reaction and how many grams are possible to be formed, right? so that's the equivalent to the theoretical yield of this reaction.

00:34 So we're going to start with one liter of octane, and we're going to have to get to grams so we can get to moles.

00:44 Right.

00:44 So the density of octane is 0 .7 .00 grams per milliliter.

00:48 So let's first convert the liters to milliliters using the prefix from an earlier chapter.

00:54 Right.

00:54 So if every thousand milliliters is one liter or a non -1.

00:59 Another way to express that is for every one milliliter, there's 10 to the minus third liters.

01:03 It's the same thing.

01:04 According to that table in the first chapter, i think it's the first chapter of this textbook.

01:09 Right, so now we're in milliliters.

01:11 So now we're going to use the density to get to grams, right? so 0 .70 grams for every one millimeter.

01:17 Then we can get to moles.

01:19 So i've calculated the weight of octane to be 114 .23 grams per mole.

01:27 So we're doing great canceling out, right? so now we're in moles of octane.

01:33 Very good.

01:34 Mill leaders cancel out.

01:35 Leaders cancel out.

01:37 So now we're going to use the conversion between the coefficients of the balanced equation.

01:42 Right.

01:42 So if you've balanced this equation before, you'll know that there are eight co2s to every one mole of octane.

01:52 So we're going to do eight moles of co2 for every one mole of octane.

02:00 And then grams of co2.

02:02 And then we're done with this part of the problem.

02:04 Right? so the molecular weight of co2 is 44 .1 grams for every one mole.

02:10 Balls cancel out...

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