Problem 116 - Chemistry The Science in Context

Question

Potassium superoxide, \(\mathrm{KO}_{2},\) reacts with carbon dioxide to form potassium carbonate and oxygen:\(4 \mathrm{KO}_{2}(s)+2 \mathrm{CO}_{2}(g) \rightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3}(s)+3 \mathrm{O}_{2}(g)\).This reaction makes potassium superoxide useful in a self-contained breathing apparatus. How much \(\mathrm{O}_{2}\) could be produced from \(2.50 \mathrm{g}\) of \(\mathrm{KO}_{2}\) and \(4.50 \mathrm{g}\) of \(\mathrm{CO}_{2} ?\)

Step-by-Step Solution

Answer

Answer: 0.845 grams of oxygen can be produced from 2.50 g of KO₂ and 4.50 g of CO₂.

Step 1: 1. Calculate molar masses

First, we need to calculate the molar masses of KO₂ and CO₂. For KO₂: - Potassium (K): 39.10 g/mol - Oxygen (O): 16.00 g/mol The molar mass of KO₂ = 39.10 g/mol + 2 * 16.00 g/mol = 71.10 g/mol For CO₂: - Carbon (C): 12.01 g/mol The molar mass of CO₂ = 12.01 g/mol + 2 * 16.00 g/mol = 44.01 g/mol

Step 2: 2. Convert masses to moles

Now, we'll convert the given masses of reactants to moles using molar masses: moles of KO₂ = (2.50 g) / (71.10 g/mol) = 0.0352 mol moles of CO₂ = (4.50 g) / (44.01 g/mol) = 0.102 mol

Step 3: 3. Determine the limiting reactant

We'll use the stoichiometry of the reaction to find the limiting reactant. The balanced chemical equation is: \(4 \mathrm{KO}_{2}(s)+2 \mathrm{CO}_{2}(g) \rightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3}(s)+3 \mathrm{O}_{2}(g)\) For every 4 moles of KO₂, 2 moles of CO₂ are required. So we'll compare the mole ratio of both reactants: Mole ratio = moles of KO₂ / moles of CO₂ = 0.0352 mol / 0.102 mol = 0.344 Since the mole ratio (0.344) is less than the stoichiometric ratio of 4:2 (or 2), KO₂ is the limiting reactant.

Step 4: 4. Calculate the moles of produced oxygen

According to the balanced chemical equation, 4 moles of KO₂ produce 3 moles of O₂. So, we'll use the moles of the limiting reactant (KO₂) to calculate the moles of produced oxygen: moles of O₂ = (3 moles O₂ / 4 moles KO₂) * moles of KO₂ = (3 O₂ / 4 KO₂) * 0.0352 mol = 0.0264 mol

Step 5: 5. Calculate the mass of produced oxygen

Finally, we'll convert the moles of produced oxygen to mass using the molar mass of O₂: - Oxygen (O): 16.00 g/mol The molar mass of O₂ = 2 * 16.00 g/mol = 32.00 g/mol mass of O₂ = moles of O₂ * molar mass of O₂ = 0.0264 mol * 32.00 g/mol = 0.845 g Thus, 0.845 grams of oxygen could be produced from 2.50 g of KO₂ and 4.50 g of CO₂.

Key Concepts

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry that allows us to convert between mass and moles, which is crucial for understanding chemical reactions. The molar mass is the mass of one mole of a substance. It is typically expressed in grams per mole (g/mol). To find this value, simply sum up the atomic masses of each element in a compound. These atomic masses can be found on the periodic table.
For potassium superoxide ( K O _2 ), we have:

• Potassium ( K ) has an atomic mass of 39.10 g/mol.
• Oxygen ( O ) has an atomic mass of 16.00 g/mol.

So, the molar mass of K O _2 is calculated as:\[ ext{Molar mass of } KO_2 = 39.10 ext{ g/mol} + 2 imes 16.00 ext{ g/mol} = 71.10 ext{ g/mol}\]Similarly, for carbon dioxide ( CO _2 ):

• Carbon ( C ) has an atomic mass of 12.01 g/mol.

Thus, the molar mass is:\[ ext{Molar mass of } CO_2 = 12.01 ext{ g/mol} + 2 imes 16.00 ext{ g/mol} = 44.01 ext{ g/mol}\]Calculating molar masses with accuracy is crucial, as these figures are used in conversions and further calculations in chemical reactions.

Stoichiometry

Stoichiometry is the study of the quantitative relationships or ratios between reactants and products in a chemical reaction. It is based on the balanced chemical equation and allows us to predict the amounts of reactants and products involved.
To determine which reactant limits the reaction, we compare the ratio of the actual moles to the required moles from the balanced equation.
Consider the reaction involving KO _2 and CO _2 :\[ 4 KO_2(s) + 2 CO_2(g) \rightarrow 2 K_2CO_3(s) + 3 O_2(g)\]We have:

• Moles of KO _2 = 0.0352 mol
• Moles of CO _2 = 0.102 mol

The stoichiometric ratio requires 4 moles of KO _2 for every 2 moles of CO _2 , which can be simplified to 2:1.
The actual mole ratio calculated is 0.344.
Since this is less than the stoichiometric ratio, KO _2 is concluded as the limiting reactant.
Understanding stoichiometry is essential for recognizing which reactant will run out first and thus determines the amount of product produced.

Chemical Reaction

A chemical reaction is a process in which substances, known as reactants, are transformed into different substances, called products. These transformations are represented by chemical equations that need to be balanced to account for the conservation of mass.
In the given reaction:\[ 4 KO_2(s) + 2 CO_2(g) \rightarrow 2 K_2CO_3(s) + 3 O_2(g)\]We observe a redox reaction where the potassium superoxide ( KO _2 ) and carbon dioxide ( CO _2 ) are used to produce potassium carbonate ( K_2CO_3 ) and oxygen gas ( O_2 ).
Each atom's count matches on both sides of the equation, reflecting a balanced reaction. This balance ensures that no atoms are created or destroyed, a principle known as the Law of Conservation of Mass.
Chemical reactions are central to chemistry and reveal how substances react and transform, making them an area of great importance and interest.