Problem 62 - Chemistry The Central Science In Si Units

Question

The reaction between potassium superoxide, \(\mathrm{KO}_{2}\), and \(\mathrm{CO}_{2}\), $$ 4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{~K}_{2} \mathrm{CO}_{3}+3 \mathrm{O}_{2} $$ is used as a source of \(\mathrm{O}_{2}\) and absorber of \(\mathrm{CO}_{2}\) in selfcontained breathing equipment used by rescue workers. (a) How many moles of \(\mathrm{O}_{2}\) are produced when \(0.400 \mathrm{~mol}\) of \(\mathrm{KO}_{2}\) reacts in this fashion? (b) How many grams of \(\mathrm{KO}_{2}\) are needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) ?

Step-by-Step Solution

Answer

(a) 0.300 mol of \(\mathrm{O}_{2}\). (b) 22.21 g of \(\mathrm{KO}_{2}\).

Step 1: Understand the Reaction

The balanced chemical equation for the reaction is:\[4 \mathrm{KO}_{2} + 2 \mathrm{CO}_{2} \rightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3} + 3 \mathrm{O}_{2}\]This equation indicates that 4 moles of \(\mathrm{KO}_{2}\) produce 3 moles of \(\mathrm{O}_{2}\). We'll use this stoichiometric relationship to solve the problem.

Step 2: Calculate Moles of \(\mathrm{O}_{2}\) from \(0.400 \mathrm{~mol}\) \(\mathrm{KO}_{2}\)

According to the equation, 4 moles of \(\mathrm{KO}_{2}\) yield 3 moles of \(\mathrm{O}_{2}\). Using this ratio:\[\frac{3 \text{ moles } \mathrm{O}_{2}}{4 \text{ moles } \mathrm{KO}_{2}} \times 0.400 \text{ moles } \mathrm{KO}_{2} = 0.300 \text{ moles } \mathrm{O}_{2}\]Therefore, 0.300 moles of \(\mathrm{O}_{2}\) are produced.

Step 3: Calculate Molar Mass of \(\mathrm{O}_{2}\)

The molar mass of \(\mathrm{O}_{2}\) is calculated by adding the atomic masses of oxygen:\[2 \times 16.00 \text{ g/mol} = 32.00 \text{ g/mol}\]

Step 4: Calculate Moles of \(\mathrm{O}_{2}\) for 7.50 g

To find how many moles correspond to 7.50 g of \(\mathrm{O}_{2}\), we use the molar mass:\[\frac{7.50 \text{ g}}{32.00 \text{ g/mol}} = 0.234375 \text{ moles }\]

Step 5: Calculate \(\mathrm{KO}_{2}\) Required for 7.50 g of \(\mathrm{O}_{2}\)

Using the stoichiometric ratio (4 moles of \(\mathrm{KO}_{2}\) produce 3 moles of \(\mathrm{O}_{2}\)):\[\frac{4 \text{ moles } \mathrm{KO}_{2}}{3 \text{ moles } \mathrm{O}_{2}} \times 0.234375 \text{ moles } \mathrm{O}_{2} = 0.3125 \text{ moles } \mathrm{KO}_{2}\]

Step 6: Calculate Molar Mass of \(\mathrm{KO}_{2}\)

The molar mass of \(\mathrm{KO}_{2}\) is calculated by summing the atomic masses: \( K = 39.10 \), \( O = 16.00 \):\[39.10 + (2 \times 16.00) = 71.10 \text{ g/mol}\]

Step 7: Calculate Grams of \(\mathrm{KO}_{2}\) Needed

To find the mass of \(\mathrm{KO}_{2}\) needed, multiply the number of moles by its molar mass:\[0.3125 \text{ moles } \times 71.10 \text{ g/mol} = 22.20625 \text{ g}\]Approximately 22.21 grams of \(\mathrm{KO}_{2}\) are required.

Key Concepts

Chemical Reactions

In a chemical reaction, substances known as reactants interact to form new substances termed products. Understanding the transformation in these reactions involves analyzing the balanced chemical equation. Each chemical equation is like a sentence that tells you what is happening in the reaction.
For instance, in the reaction involving potassium superoxide (\(\mathrm{KO}_{2}\)) and carbon dioxide (\(\mathrm{CO}_{2}\)), the balanced equation is:\[4 \mathrm{KO}_{2} + 2 \mathrm{CO}_{2} \rightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3} + 3 \mathrm{O}_{2}\]This equation communicates several crucial pieces of information:

• It tells us the specific number of molecules or moles of each reactant and product involved in the reaction.
• It is also balanced, meaning atoms on both sides of the equation are equal, adhering to the law of conservation of mass.
• For every 4 moles of \(\mathrm{KO}_{2}\), 2 moles of \(\mathrm{CO}_{2}\) react to produce 2 moles of \(\mathrm{K}_{2}\mathrm{CO}_{3}\) and 3 moles of \(\mathrm{O}_{2}\).

We can extract important information on how the substances interact and what they yield. By doing so, we dive into the essentials of stoichiometry, which will guide us through these calculations.

Mole Calculations

Mole calculations are central to understanding chemical reactions, as they allow chemists to work with the macroscopic amounts of substances involved. A mole is a unit that chemists use to express amounts of a chemical substance. One mole is equivalent to Avogadro's number, \(6.022 \times 10^{23}\) particles, atoms, or molecules.
Applying this concept to our exercise, we focus on using moles to quantify the amount of substance produced or reacted. Let's say you have \(0.400\) moles of \(\mathrm{KO}_{2}\):

• From the balanced equation, we know 4 moles of \(\mathrm{KO}_{2}\) can produce 3 moles of \(\mathrm{O}_{2}\).
• To find out how many moles of \(\mathrm{O}_{2}\) are produced, we use the ratio \(\frac{3 \text{ moles } \mathrm{O}_{2}}{4 \text{ moles } \mathrm{KO}_{2}} \).
• By multiplying this ratio by \(0.400 \) moles \(\mathrm{KO}_{2}\), we compute that \(0.300\) moles of \(\mathrm{O}_{2}\) are produced.

Remember, determining how much of a product is formed or reactant is needed involves these ratio-based calculations, highlighting why mastering mole calculations is key in chemistry.

Oxygen Production

Oxygen production in this reaction is noteworthy, especially in applications like self-contained breathing equipment. In this scenario, rescue workers gain breathable \(\mathrm{O}_{2}\) using controlled chemical reactions under emergency conditions.
The chemical reaction between \(\mathrm{KO}_{2}\) and \(\mathrm{CO}_{2}\) not only cleanses the air of carbon dioxide but also ensures a supply of oxygen. To understand how much \(\mathrm{KO}_{2}\) is needed to produce a certain mass of \(\mathrm{O}_{2}\), let's consider a situation where you need \(7.50 \text{ g}\) of \(\mathrm{O}_{2}\):

• First, calculate how many moles are in \(7.50 \text{ g}\) of \(\mathrm{O}_{2}\): Using its molar mass \(32.00\, \text{g/mol}\), \(\frac{7.50 \text{ g}}{32.00 \text{ g/mol}} = 0.234375 \text{ moles of } \mathrm{O}_{2}\).
• Using the stoichiometric ratio, determine the moles of \(\mathrm{KO}_{2}\) needed: \(\frac{4 \text { moles } \mathrm{KO}_{2}}{3 \text{ moles } \mathrm{O}_{2}} \times 0.234375 \text{ moles } \mathrm{O}_{2} = 0.3125 \text{ moles of } \mathrm{KO}_{2}\).
• Finally, from the molar mass of \(\mathrm{KO}_{2}\) \((71.10 \text{ g/mol})\), calculate the needed mass: \(0.3125 \text{ moles} \times 71.10 \text{ g/mol} = 22.21 \text{ g of } \mathrm{KO}_{2}\).

Such calculations enable efficient planning and execution of oxygen production in practical scenarios.