Problem 62 - Chemistry The Central Science

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 self-contained 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 \(\mathrm{O}_{2} ?\) (c) How many grams of \(\mathrm{CO}_{2}\) are used when 7.50 \(\mathrm{g}\) of \(\mathrm{O}_{2}\) are produced?

Step-by-Step Solution

Answer

(a) When 0.400 mol of KO₂ reacts, 0.300 mol of O₂ is produced. (b) 17.24 g of KO₂ is needed to form 7.50 g of O₂. (c) 7.00 g of CO₂ is used when 7.50 g of O₂ is produced.

Step 1: (a) Calculate moles of O₂ produced

To calculate the moles of O₂ produced when 0.400 mol of KO₂ reacts, we can use the stoichiometry of the balanced chemical equation. From the balanced equation, we see that 4 moles of KO₂ produce 3 moles of O₂. So, for 0.400 mol of KO₂, Moles of O₂ = (Moles of KO₂ × Moles of O₂ produced) / (Moles of KO₂ reacting) Moles of O₂ = (0.400 mol × 3) / 4 Moles of O₂ produced = 0.300 mol

Step 2: (b) Calculate grams of KO₂ needed to produce 7.50 g O₂

First, we need to convert the given mass of O₂ to moles: Moles of O₂ = mass of O₂ / molar mass of O₂ Moles of O₂ = 7.50 g / 32.00 g/mol Moles of O₂ = 0.2344 mol Now, we find how many moles of KO₂ are needed to produce 0.2344 moles of O₂. From the balanced equation, 4 moles of KO₂ produce 3 moles of O₂. So, Moles of KO₂ = (Moles of O₂ × Moles of KO₂ reacting) / Moles of O₂ produced Moles of KO₂ = (0.2344 mol × 4) / 3 Moles of KO₂ = 0.3125 mol Now, we convert moles of KO₂ to grams: Mass of KO₂ = moles of KO₂ × molar mass of KO₂ Mass of KO₂ = 0.3125 mol × (39.10 g/mol + 16.00 g/mol) Mass of KO₂ = 17.24 g

Step 3: (c) Calculate grams of CO₂ used to produce 7.50 g O₂

We know that 0.2344 mol of O₂ is produced, which we found in part (b). We will now find the moles of CO₂ used in the reaction. From the balanced equation, 4 moles of KO₂ react with 2 moles of CO₂ to produce 3 moles of O₂. Moles of CO₂ = (Moles of O₂ × Moles of CO₂ reacting) / Moles of O₂ produced Moles of CO₂ = (0.2344 mol × 2) / 3 Moles of CO₂ = 0.1563 mol Now, we convert moles of CO₂ to grams: Mass of CO₂ = moles of CO₂ × molar mass of CO₂ Mass of CO₂ = 0.1563 mol × (12.01 g/mol + 2 * 16.00 g/mol) Mass of CO₂ = 7.00 g

Key Concepts

Understanding Chemical Reactions

Chemical reactions occur when substances interact to form new products. In this exercise, the reaction involves potassium superoxide (\(\mathrm {KO}_2\)) and carbon dioxide (\(\mathrm{CO}_2\)). This specific reaction is vital as it acts both as a source of oxygen (\(\mathrm{O}_2\)) and helps in absorbing carbon dioxide. When rescue workers require self-contained breathing equipment, such reactions ensure a reliable supply of oxygen while managing carbon dioxide levels. Every chemical reaction involves the breaking of old bonds and the forming of new bonds, leading to different substances forming as products from reactants. Being aware of the roles of reactants and products in a chemical reaction allows understanding the functionality behind reactions, such as this one, utilized for creating breathable environments.

Molar Mass Calculation

Molar mass serves as a conversion factor between grams and moles for a given substance. It is essentially the "weight" of one mole of a substance, where one mole equals Avogadro's number (\(6.022 \times 10^{23}\)) of particles. In this task, calculating the molar mass of oxygen (\(\mathrm{O}_2\)), potassium superoxide (\(\mathrm{KO}_2\)), and carbon dioxide (\(\mathrm{CO}_2\)) was crucial. - For \(\mathrm{O}_2\), each molecule has a molar mass of 32.00 g/mol.- Potassium superoxide, \(\mathrm{KO}_2\), combines the atomic masses of potassium (39.10 g/mol) and two oxygen atoms (2 × 16.00 g/mol), resulting in 71.10 g/mol.- Carbon dioxide's molar mass is calculated by adding the atomic masses of one carbon (12.01 g/mol) and two oxygen atoms, yielding 44.01 g/mol.Understanding molar mass is integral to converting between the mass of a substance and its amount in moles—a process that allows us to predict and measure quantities in chemical reactions accurately.

Balanced Chemical Equations

Balanced chemical equations are foundational to understanding stoichiometry. They ensure that the same number of each type of atom is present on both sides of the equation. In our exercise, the chemical equation is \[4 \mathrm{KO}_2 + 2 \mathrm{CO}_2 \longrightarrow 2 \mathrm{K}_2\mathrm{CO}_3 + 3 \mathrm{O}_2 \]This equation is balanced, meaning four moles of \(\mathrm{KO}_2\) react with two moles of \(\mathrm{CO}_2\) to produce three moles of \(\mathrm{O}_2\) and two moles of \(\mathrm{K}_2\mathrm{CO}_3\). Balanced equations communicate how molecules and atoms rearrange and are essential for stoichiometric calculations which predict product and reactant relations in reactions.They provide the conversion factors necessary for determining mole ratios of each reactant and product, allowing for precise calculation of substance requirements whether producing a specific amount of product or converting back to reactants. Understanding these ratios is key to solving problems regarding how much of a substance is needed or produced in chemical reactions.