Problem 122 - Introductory Chemistry Atoms First

Question

Consider the unbalanced chemical equation \(\mathrm{KCl}+\mathrm{O}_{2} \rightarrow \mathrm{KClO}_{3}\) If you react \(42.6 \mathrm{~g}\) of \(\mathrm{KCl}\) and \(36.5 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and the reaction has a percent yield of \(56.0 \%\), how many grams of \(\mathrm{KClO}_{3}\) are produced?

Step-by-Step Solution

Answer

Considering a percent yield of 56.0%, approximately 39.1 grams of potassium chlorate (KClO3) will be produced.

Step 1: Write the balanced chemical equation

First, we need to balance the given chemical equation: 2 KCl + 3 O2 -> 2 KClO3

Step 2: Calculate the moles of reactants (KCl and O2)

Next, we will convert the given masses of KCl and O2 into moles. For this, we'll need their molar masses: Molar mass of KCl \(= 39.10~g/mol~(K) + 35.45~g/mol~(Cl) = 74.55~g/mol\) Molar mass of O2 \(= 2 × 16.00~g/mol~(O) = 32.00~g/mol\) Moles of KCl \(= \frac{42.6~g}{74.55~g/mol} = 0.571~mol\) Moles of O2 \(= \frac{36.5~g}{32.00~g/mol} = 1.14~mol\)

Step 3: Determine the limiting reactant

To identify the limiting reactant, we will compare the mole-to-coefficient ratios of the reactants: Mole-to-coefficient ratio of KCl \(= \frac{0.571~mol}{2} = 0.285\) Mole-to-coefficient ratio of O2 \(= \frac{1.14~mol}{3} = 0.380\) Since the mole-to-coefficient ratio of KCl is smaller, KCl is the limiting reactant.

Step 4: Calculate the theoretical yield of KClO3 in grams

We will use stoichiometry to find the theoretical yield of KClO3 based on the limiting reactant (KCl). The molar mass of KClO3 is \(= 39.10~g/mol~(K) + 35.45~g/mol~(Cl) + 3 × 16.00~g/mol~(O) = 122.55~g/mol\) For every 2 moles of KCl, 2 moles of KClO3 are produced (according to the balanced chemical equation). Therefore, we can find the moles of KClO3 produced as \(0.571~mol~(KCl) × \frac{2~mol~(KClO3)}{2~mol~(KCl)} = 0.571~mol~(KClO3)\) The theoretical yield of KClO3 in grams is: \(0.571~mol~(KClO3) × \frac{122.55~g/mol~(KClO3)}{1~mol~(KClO3)} = 69.9~g~(KClO3)\)

Step 5: Calculate the actual yield of KClO3 considering the given percent yield

The actual yield can be calculated using the equation: Actual yield \(= \frac{Percent~Yield}{100} × Theoretical~Yield\) Actual yield of KClO3 \(= \frac{56.0\%}{100} × 69.9~g = 39.1~g\) Considering a percent yield of 56.0%, approximately 39.1 grams of potassium chlorate (KClO3) will be produced.

Key Concepts

Limiting reactant

When you're working with chemical reactions, the limiting reactant is the substance that gets completely used up first. Once this reactant is all used up, the reaction can't keep going. It's a crucial piece of the puzzle because it determines how much product you can make. In the exercise above, we calculated the moles of both potassium chloride (KCl) and oxygen ( O_2 ) to figure out which one runs out first.
To identify the limiting reactant, we looked at the mole-to-coefficient ratios from the balanced equation:

• KCl has a mole-to-coefficient ratio of 0.285.
• Oxygen's ratio is 0.380.

This tells us that KCl is the limiting reactant because its ratio is lower. It's essential to always start with a balanced equation to make sure our comparisons are accurate.

Percent yield

Percent yield is a delightful measure of an experiment's 'success'. It tells you how close you got to what you were theoretically supposed to get. Nobody likes waste, right? Understanding percent yield helps us see how efficient our reactions are.
It's calculated using the formula:

• Actual yield / Theoretical yield × 100%

In this case, our theoretical yield was 69.9 grams of KClO_3. However, in real life, we often don't get this exact amount. Here, the reaction had a percent yield of 56.0%.
This means that when we apply it to the theoretical yield: \[ \text{Actual yield} = \frac{56.0}{100} \times 69.9~g = 39.1~g \]This discrepancy between theoretical and actual yields is expected due to losses like side reactions or incomplete reactions.

Molar mass calculation

Molar mass is fundamental when it comes to converting between grams and moles, a necessary step in many stoichiometry problems. It's like the bridge that connects mass and moles. The molar mass is simply the mass of one mole of a substance, usually measured in grams per mole.
For this exercise, we calculated the molar masses for KCl and O_2 :

• Molar mass of KCl is 74.55 g/mol.
• Molar mass of O_2 is 32.00 g/mol.

These calculations involve adding up the atomic masses of each element in the compound, as you'd find on the periodic table.
We used these values to find the number of moles of the reactants, setting the stage for figuring out the limiting reactant and the theoretical yield. Remember, molar mass is your ticket to moving seamlessly between the measured weight of compounds and the number of particles involved!