Chapter 3: Problem 75
When potassium chlorate is burned, potassium chloride and oxygen are formed. (a) Write a balanced equation for the reaction. (b) How much potassium chlorate must be burned to produce \(198.5 \mathrm{~g}\) of oxygen? The yield is found to be \(83.2 \%\).
Short Answer
Expert verified
Answer: 609.25 g of potassium chlorate.
Step by step solution
01
Write the balanced equation
First, write down the unbalanced equation for the given reaction:
$$
\mathrm{KClO_3} \rightarrow \mathrm{KCl} + \mathrm{O_2}
$$
To balance the equation, we need to adjust the coefficients in the reaction to ensure that the number of atoms of each element is the same on both sides.
Balanced equation:
$$
\mathrm{2KClO_3} \rightarrow \mathrm{2KCl} + \mathrm{3O_2}
$$
02
Calculate moles of oxygen
Now that we have a balanced equation, we can calculate the moles of oxygen needed. We are given that we need \(198.5 \mathrm{~g}\) of oxygen and we know the molar mass of oxygen is approximately \(32 \mathrm{~g/mol}\).
Number of moles of oxygen = \(\frac{\text{mass}}{\text{molar mass}}\)
$$
\frac{198.5 \mathrm{~g}}{32 \mathrm{~g/mol}} = 6.203125 \mathrm{~mol}
$$
03
Calculate moles of potassium chlorate
Now we need to find how many moles of potassium chlorate are needed to produce the desired amount of oxygen, according to the balanced equation. From the balanced equation, we can see that 2 moles of potassium chlorate produce 3 moles of oxygen.
$$
\frac{\text{moles of potassium chlorate}}{\text{moles of oxygen}} = \frac{2}{3}
$$
To calculate the moles of potassium chlorate required, simply multiply the moles of oxygen by the coefficient ratio.
$$
\text{moles of potassium chlorate} = 6.203125 \mathrm{~mol} \times \frac{2}{3} = 4.13541667 \mathrm{~mol}
$$
04
Calculate grams of potassium chlorate
To find the mass of potassium chlorate required, we need to consider the yield given (83.2%). We will multiply the calculated moles of potassium chlorate by its molar mass and then divide by the yield to obtain the actual amount needed.
Molar mass of potassium chlorate:
$$
\mathrm{KClO_3} = 39.1 (\mathrm{K}) + 35.5 (\mathrm{Cl}) + 3 \times 16 (\mathrm{O}) = 122.6 \mathrm{~g/mol}
$$
Mass of potassium chlorate required (without considering yield):
$$
4.13541667 \mathrm{~mol} \times 122.6 \mathrm{~g/mol} = 507.073100685 \mathrm{~g}
$$
Taking into account the yield (83.2%):
$$
\frac{507.073100685 \mathrm{~g}}{0.832} = 609.24897156 \mathrm{~g}
$$
05
Final answer
Thus, \(609.25 \mathrm{~g}\) (rounded to 2 decimal places) of potassium chlorate must be burned to produce \(198.5 \mathrm{~g}\) of oxygen with a yield of \(83.2 \%\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Stoichiometry
Stoichiometry is the mathematical relationship between reactants and products in a chemical reaction. It's like a recipe for a reaction! When you bake a cake, you follow a recipe with specific amounts of ingredients to get the perfect cake. In chemistry, stoichiometry helps you use the right amounts of reactants to produce desired products.
Let's take a look at our problem with potassium chlorate. Here, stoichiometry helps determine how much potassium chlorate we need to produce a certain amount of oxygen. The balanced chemical equation is the key to stoichiometry, ensuring that both sides of the reaction have the same number of atoms of each element. This equation guides you to find the precise relationship between the amounts of reactants and products. By using stoichiometry, we make sure we don’t use too much or too little of the substances involved in the reaction!
To calculate use:
Let's take a look at our problem with potassium chlorate. Here, stoichiometry helps determine how much potassium chlorate we need to produce a certain amount of oxygen. The balanced chemical equation is the key to stoichiometry, ensuring that both sides of the reaction have the same number of atoms of each element. This equation guides you to find the precise relationship between the amounts of reactants and products. By using stoichiometry, we make sure we don’t use too much or too little of the substances involved in the reaction!
To calculate use:
- The balanced chemical equation
- Mole ratios between reactants and products
- Molar masses for conversions
Yield Calculation
Yield calculation is about understanding how much of a product is actually obtained from a reaction, compared to what could potentially be made. The yield is expressed as a percentage; this is the percentage of the theoretical yield that is actually obtained.
In our exercise, the yield of oxygen from burning potassium chlorate is 83.2%. This means that, due to losses or inefficiencies, only 83.2% of the expected product is actually realized. To find out how much potassium chlorate is required to produce 198.5 g of oxygen when taking yield into account, adjust the theoretical calculation using this percentage. Let's put it into practice:
In our exercise, the yield of oxygen from burning potassium chlorate is 83.2%. This means that, due to losses or inefficiencies, only 83.2% of the expected product is actually realized. To find out how much potassium chlorate is required to produce 198.5 g of oxygen when taking yield into account, adjust the theoretical calculation using this percentage. Let's put it into practice:
- Calculate the theoretical required amount, assuming 100% yield
- Divide the theoretical amount by the yield percentage (converted to a decimal)
Molar Mass Calculation
Molar mass is crucial for converting between grams and moles, which are the units chemists commonly use to describe amounts of substances. The molar mass of a compound is the mass of one mole of that compound, and it is usually expressed in grams per mole (\(\text{g/mol}\)). You can find the molar mass by adding up the atomic masses of all the atoms in the compound.
For example, the molar mass of potassium chlorate (\(\mathrm{KClO_3}\)) is calculated by adding:
For example, the molar mass of potassium chlorate (\(\mathrm{KClO_3}\)) is calculated by adding:
- Potassium (\(\text{K}\)) = 39.1 \(\text{g/mol}\)
- Chlorine (\(\text{Cl}\)) = 35.5 \(\text{g/mol}\)
- Oxygen (\(\text{O}\)) = 3 x 16 \(\text{g/mol}\)
Balancing Chemical Equations
Balancing chemical equations is crucial because it reflects the law of conservation of mass: matter cannot be created or destroyed. This means the number of atoms for each element must be the same on both sides of the reaction.
In the problem, the unbalanced reaction equation for burning potassium chlorate is:\[\mathrm{KClO_3} \rightarrow \mathrm{KCl} + \mathrm{O_2}\]
To balance the equation, follow these steps:
This equation now has the same number of potassium (K), chlorine (Cl), and oxygen (O) atoms on both sides, making it balanced. Balancing helps you understand the exact proportions needed for a reaction to occur correctly, which is essential for accurate stoichiometric calculations.
In the problem, the unbalanced reaction equation for burning potassium chlorate is:\[\mathrm{KClO_3} \rightarrow \mathrm{KCl} + \mathrm{O_2}\]
To balance the equation, follow these steps:
- Write down the number of atoms for each element on both sides
- Add coefficients to balance the atoms one element at a time
- Start by balancing single elements, then move to compounds that contain multiple atoms of different elements
This equation now has the same number of potassium (K), chlorine (Cl), and oxygen (O) atoms on both sides, making it balanced. Balancing helps you understand the exact proportions needed for a reaction to occur correctly, which is essential for accurate stoichiometric calculations.