The citizens of the world burn the fossil fuel equivalent of 7 ×10^12 kg of petroleum per year.

step 1 Okay, again, we get a multi -part question here. So, first of all, we've got a balanced chemical

The citizens of the world burn the fossil fuel equivalent of...

The citizens of the world burn the fossil fuel equivalent of $7 \times 10^{12} \mathrm{~kg}$ of petroleum per year. Assume that all of this petroleum is in the form of octane $\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)$ and calculate how much $\mathrm{CO}_{2}$ (in kg) the world produces from fossil fuel combustion per year. (Hint: Begin by writing a balanced equation for the combustion of octane.) If the atmosphere currently contains approximately $3 \times 10^{15} \mathrm{~kg}$ of $\mathrm{CO}_{2}$, how long will it take for the world's fossil fuel combustion to double the amount of atmo- spheric carbon dioxide?

Key Concepts

Rate and Accumulation Calculations

Rate and accumulation calculations involve determining how the production or consumption of a substance over time leads to overall changes in a system. In environmental chemistry, this type of calculation is used to project how continuous emissions, such as CO2 from fossil fuel combustion, accumulate in the atmosphere over time, thereby helping assess long-term impacts on atmospheric composition.

Mass-to-Mole and Mole-to-Mass Conversions

Converting mass to moles, and vice versa, is crucial in chemical calculations. These conversions utilize the substance's molar mass to interconvert between grams (or kilograms) and moles, allowing for the application of the mole ratios from the balanced equation to determine the mass of products formed from a given mass of reactants.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It uses the molar ratios derived from a balanced chemical equation to convert between amounts of different substances, such as from the mass of fuel consumed to the mass of CO2 produced in a combustion reaction.

Mole Concept and Molar Mass Calculations

The mole concept provides a bridge between the mass of a substance and the number of atoms or molecules it contains using Avogadro’s number. Molar mass is the mass of one mole of a substance and is used to convert between the mass of a chemical and its number of moles, which is key in applying stoichiometric ratios in chemical calculations.

Balanced Chemical Equations

A balanced chemical equation represents the conservation of mass by showing that the number of atoms of each element is the same on both the reactant and product sides of a reaction. In combustion processes, creating a balanced equation allows for the accurate determination of stoichiometric ratios, which are essential for calculating how much product (such as CO2) is produced from a given amount of reactant (such as fuel).

Transcript

00:01 Okay, again, we get a multi -part question here.

00:04 So, first of all, we've got a balanced chemical equation here for octane.

00:09 And if we can burn seven times 10 of the 12 kilograms of octane every single year, how many kilograms of carbon dioxide does that equate to? so this is a simple stochometry problem.

00:25 So we start with the setup.

00:28 We start with what's given to us.

00:29 So seven times 10 of the 12 kilograms of octane.

00:35 In order to switch substances, you know, when we're comparing in a balanced equation, first of all, we need to convert to moles.

00:43 So i'm going to use the molar mass here.

00:45 If i have eight carbons and 18 hydrogens, the molar mass is 114 .26 grams for every one mole of octane.

00:56 Okay, so you always have to convert to moles first.

01:01 Okay, and we do that so that we can then do a mole -to -mole ratio.

01:04 Here's where the balance chemical equation comes in.

01:07 So there's two moles of octane for every 16 moles of carbon dioxide.

01:14 And then since i want to know something about the mass of carbon dioxide, i then go from moles to grams.

01:23 So for every one mole of carbon dioxide, there are 44 .01 grams of carbon.

01:31 Now notice when i was canceling units, i canceled the grams here, but left the prefix kilo...

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