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Chapter 3: Problem 83

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 \(\mathrm{O}_{2}\).)

Short Answer

Expert verified
18.0 grams of \(\mathrm{O}_2\) can be obtained.

Step by step solution

01

Write the Balanced Chemical Equation

The thermal decomposition of potassium chlorate (\(\mathrm{KClO}_3\)) is described by the following chemical reaction:\[2\mathrm{KClO}_3 \rightarrow 2\mathrm{KCl} + 3\mathrm{O}_2\]This balanced equation tells us that 2 moles of \(\mathrm{KClO}_3\) yield 3 moles of \(\mathrm{O}_2\).
02

Calculate Molar Mass of KClO鈧�

To proceed with the calculation, we need the molar mass of \(\mathrm{KClO}_3\):- Potassium (K): 39.1 g/mol- Chlorine (Cl): 35.5 g/mol- Oxygen (O): 16.0 g/mol (Note: there are three atoms of oxygen)So, the molar mass of \(\mathrm{KClO}_3\) is given by: \[\mathrm{Molar\ Mass\ of\ KClO}_3 = 39.1 + 35.5 + 3(16.0) = 122.6\ \mathrm{g/mol}\]
03

Determine Moles of KClO鈧�

Given mass of \(\mathrm{KClO}_3\) is 46.0 g. We use the molar mass to find the number of moles:\[\text{Moles of } \mathrm{KClO}_3 = \frac{46.0}{122.6} \approx 0.375\ \mathrm{moles}\]
04

Use Stoichiometry to Find Moles of O鈧� Produced

From the balanced chemical equation, 2 moles of \(\mathrm{KClO}_3\) produce 3 moles of \(\mathrm{O}_2\). Thus, the moles of \(\mathrm{O}_2\) produced are:\[\text{Moles of } \mathrm{O}_2 = 0.375 \times \frac{3}{2} = 0.5625\ \mathrm{moles}\]
05

Calculate Mass of O鈧� Produced

The molar mass of \(\mathrm{O}_2\) (oxygen gas) is:- Oxygen (O) = 16.0 g/mol, thus \(\mathrm{O}_2 = 2 \times 16.0 = 32.0\ \mathrm{g/mol}\)Now, calculate the mass:\[\text{Mass of } \mathrm{O}_2 = 0.5625 \times 32.0 = 18.0\ \mathrm{g}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Decomposition
In chemistry, thermal decomposition is a process where a chemical compound breaks down into simpler substances when heated. This process is a type of chemical reaction characterized by the compounds separating due to heat. One common example is the decomposition of potassium chlorate (\(\mathrm{KClO}_3\)). When potassium chlorate is heated, it breaks down into potassium chloride (\(\mathrm{KCl}\)) and oxygen gas (\(\mathrm{O}_2\)).
In a laboratory setting, this reaction is useful for preparing oxygen gas.
  • During the reaction, heat provides energy to break the chemical bonds in the compound.
  • This type of reaction is often endothermic, meaning it absorbs heat from its surroundings.
  • Safety is key, as the production of gases can cause pressure build-up.
The ability to prepare oxygen through thermal decomposition demonstrates practical applications of chemical principles.
Potassium Chlorate
Potassium chlorate, \(\mathrm{KClO}_3\), is a chemical compound that plays a significant role in various applications. It's composed of potassium (K), chlorine (Cl), and oxygen (O). This compound is renowned for its oxidizing abilities and is widely used in the laboratory to produce oxygen gas through thermal decomposition.
In the context of decomposition, potassium chlorate serves as a source of oxygen: \[ 2\mathrm{KClO}_3 \rightarrow 2\mathrm{KCl} + 3\mathrm{O}_2 \]
Here are some key points about potassium chlorate:
  • It is commonly found as a white crystalline solid.
  • It has oxidizing properties which make it useful in fireworks, explosives, and safety matches.
  • In a controlled environment, potassium chlorate is heated to mediate the release of oxygen, which is essential for various experimental setups and demonstrations.
Understanding the properties and uses of potassium chlorate allows chemists to safely and effectively employ it in practical applications.
Stoichiometry
Stoichiometry is an essential concept that involves the calculation of reactants and products in chemical reactions. It is based on the principle that matter is conserved in a chemical reaction, meaning the mass of reactants equals the mass of products.
In the thermal decomposition of potassium chlorate, stoichiometry helps us determine how much oxygen is produced from a given amount of \(\mathrm{KClO}_3\).
  • Stoichiometry uses balanced chemical equations as a reference for calculating quantities.
  • It involves a series of steps to convert between different units, like grams and moles, to predict amounts of products formed.
  • Understanding stoichiometry allows for making precise predictions about the outcomes of reactions, which is crucial in laboratory settings.
Practical application of stoichiometry ensures the efficient use of chemicals and helps prevent waste and hazards in experiments.
Balanced Chemical Equation
A balanced chemical equation is a representation of a chemical reaction that shows the relationship between reactants and products with equal numbers of atoms for each element on both sides of the equation.
For the decomposition of \(\mathrm{KClO}_3\), the balanced equation is:\[2\mathrm{KClO}_3 \rightarrow 2\mathrm{KCl} + 3\mathrm{O}_2\]
This equation is balanced because:
  • There are 2 potassium (K), 2 chlorine (Cl), and 6 oxygen (O) atoms on both the reactant and product sides.
  • It reflects the conservation of mass, a fundamental principle of chemistry.
  • Balancing equations ensures mathematical accuracy in stoichiometry, helping in the calculation of the amounts of substances involved in the reaction.
Balanced equations are crucial as they provide the necessary information to predict the amounts of products that will be generated from given reactants.

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Most popular questions from this chapter

Ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\), an important industrial organic chemical, can be prepared by heating hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) at \(800^{\circ} \mathrm{C}\) : $$ \mathrm{C}_{6} \mathrm{H}_{14} \stackrel{\Delta}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{4}+\text { other products } $$ If the yield of ethylene production is 42.5 percent, what mass of hexane must be used to produce \(481 \mathrm{~g}\) of ethylene?

Lactic acid, which consists of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O},\) has long been thought to be responsible for muscle soreness following strenuous exercise. Determine the empirical formula of lactic acid given that combustion of a 10.0 -g sample produces \(14.7 \mathrm{~g} \mathrm{CO}_{2}\) and \(6.00 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\).

Octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is a component of gasoline. Complete combustion of octane yields \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\). Incomplete combustion produces \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO},\) which not only reduces the efficiency of the engine using the fuel but is also toxic. In a certain test run, 1.000 gallon (gal) of octane is burned in an engine. The total mass of \(\mathrm{CO}, \mathrm{CO}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\) produced is \(11.53 \mathrm{~kg} .\) Calculate the efficiency of the process; that is, calculate the fraction of octane converted to \(\mathrm{CO}_{2}\). The density of octane is \(2.650 \mathrm{~kg} / \mathrm{gal}\).

In combustion analysis, is the combined mass of the products \(\left(\mathrm{CO}_{2}\right.\) and \(\mathrm{H}_{2} \mathrm{O}\) ) less than, equal to, or greater than the combined mass of the compound that is combusted and the \(\mathrm{O}_{2}\) that reacts with it? Explain.

Give an everyday example that illustrates the limiting reactant concept.
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