A common laboratory preparation of oxygen gas is the thermal...

A common laboratory preparation of oxygen gas is the thermal decomposition of potassium chlorate $\left(\mathrm{KClO}_{3}\right)$. Assuming complete decomposition, calculate the number of grams of $\mathrm{O}_{2}$ gas that can be obtained from $46.0 \mathrm{~g}$ of $\mathrm{KClO}_{3}$. (The products are $\mathrm{KCl}$ and $\left.\mathrm{O}_{2} .\right)$

Key Concepts

Thermal Decomposition Reaction

This is a type of chemical reaction where a single compound breaks down into two or more simpler substances when heated. The process is governed by heat energy, which provides the activation energy needed to break chemical bonds, resulting in the formation of products that are typically more stable at higher temperatures.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. It uses the balanced chemical equation to convert between grams, moles, and numbers of particles, helping to determine the exact amount of each substance involved in the reaction.

Mole Concept and Molar Mass

The mole concept is a fundamental principle in chemistry used to count quantities of particles, defining one mole as 6.022 x 10^23 entities. Molar mass connects the mass of a substance to the number of moles, allowing the conversion between the mass of a chemical sample and the number of moles, which is crucial for calculating proportions in chemical reactions.

Balanced Chemical Equation

A balanced chemical equation accurately represents the conservation of mass by ensuring that the number of atoms for each element is the same on both the reactant and product sides. It provides the stoichiometric coefficients necessary for determining the molar ratios between substances, which is essential for any quantitative analysis in chemical reactions.

Transcript

00:01 So for this problem, we have the thermal decomposition of potassium chloride into kcl and oxygen gas, and we want to find how many grams of o2 gas can be obtained from some amount of kcl -3.

00:14 So the first thing that we need to do is we need to actually write this equation out.

00:18 So we know that our only reactant is k -c -o -3, and that our products are k -cl and o2.

00:29 So the first thing that we need to do is we need to balance this obviously.

00:34 So we see potassium balanced.

00:38 We have one on each side.

00:40 Then we need to look to the chlorine.

00:42 The chlorine is also balanced.

00:44 We have one on each side.

00:47 The problem here is the oxygen.

00:51 Now we have three on the left hand side and only two on the right hand side.

00:56 So in order to make three on each side, i just put a three over two in front of the but i can't have fractions in balanced equations.

01:04 So i multiply this whole thing by 2.

01:06 And i get that my final balanced equation is going to be 2kcl -3 yields kcl plus 2kcl plus 3 -02.

01:19 So that is my balanced equation.

01:21 Now if i'm going to convert grams of kcl -03 into grams of 02, i'm going to have to find the molar masses of both o2 and kclo3.

01:33 So it's first to kclo3...

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