Problem 115 - Chemistry The Central Science In Si Units

Question

Potassium superoxide, \(\mathrm{KO}_{2},\) is often used in oxygen masks (such as those used by firefighters) because \(\mathrm{KO}_{2}\) reacts with \(\mathrm{CO}_{2}\) to release molecular oxygen. Experiments indicate that 2 mol of \(\mathrm{KO}_{2}(s)\) react with each mole of \(\mathrm{CO}_{2}(g) .(\mathbf{a})\) The products of the reaction are \(\mathrm{K}_{2} \mathrm{CO}_{3}(s)\) and \(\mathrm{O}_{2}(g) .\) Write a balanced equation for the reaction between \(\mathrm{KO}_{2}(s)\) and \(\mathrm{CO}_{2}(g) .(\mathbf{b})\) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced? (c) What mass of \(\mathrm{KO}_{2}(s)\) is needed to consume \(18.0 \mathrm{~g} \mathrm{CO}_{2}(g)\) ? What mass of \(\mathrm{O}_{2}(g)\) is produced during this reaction?

Step-by-Step Solution

Answer

To consume 18.0 g of \(\mathrm{CO}_2\), 58.3 g of \(\mathrm{KO}_2\) is required, producing 19.6 g of \(\mathrm{O}_2\).

Step 1: Write the unbalanced chemical equation

Based on the description given, the reactants are potassium superoxide, \(\mathrm{KO}_2\), and carbon dioxide, \(\mathrm{CO}_2\). The products that are expected from the reaction are potassium carbonate, \(\mathrm{K}_2\mathrm{CO}_3\), and oxygen gas, \(\mathrm{O}_2\). Write the skeletal equation: \(\mathrm{KO}_2(s) + \mathrm{CO}_2(g) \rightarrow \mathrm{K}_2\mathrm{CO}_3(s) + \mathrm{O}_2(g)\).

Step 2: Balance the chemical equation

The experiments indicate 2 moles of \(\mathrm{KO}_2\) react with 1 mole of \(\mathrm{CO}_2\). Therefore, the balanced equation is:\[2 \mathrm{KO}_2(s) + \mathrm{CO}_2(g) \rightarrow \mathrm{K}_2\mathrm{CO}_3(s) + \frac{3}{2} \mathrm{O}_2(g)\]. Adjust the equation to use full integer coefficients by multiplying through by 2: \[4 \mathrm{KO}_2(s) + 2 \mathrm{CO}_2(g) \rightarrow 2 \mathrm{K}_2\mathrm{CO}_3(s) + 3 \mathrm{O}_2(g)\].

Step 3: Determine oxidation numbers

For \(\mathrm{KO}_2\): K is +1, O2^(-1/2) (peroxide ion). For \(\mathrm{CO}_2\): C is +4, O is -2. In \(\mathrm{K}_2\mathrm{CO}_3\): K is +1, C is +4, and O is -2. In \(\mathrm{O}_2\): O is 0. Comparing oxidation states shows that no element in this reaction changes its oxidation state; hence, there is neither reduction nor oxidation taking place.

Step 4: Determine mass of \(\mathrm{KO}_2\) needed

Molar mass of \(\mathrm{CO}_2\) is calculated as follows: Carbon (12 g/mol) + two Oxygens (2*16 g/mol) = 44 g/mol. Given 18.0 g of \(\mathrm{CO}_2\), calculate moles of \(\mathrm{CO}_2\): \(\text{Moles of } \mathrm{CO}_2 = \frac{18}{44} = 0.409\text{ moles.}\)According to the balanced equation, 2 moles of \(\mathrm{KO}_2\) are needed per mole of \(\mathrm{CO}_2\), so 0.819 moles of \(\mathrm{KO}_2\) are needed. Molar mass of \(\mathrm{KO}_2\) is: Potassium (39.1 g/mol) + two Oxygens (2*16 g/mol) = 71.1 g/mol. Mass of \(\mathrm{KO}_2\) needed = \(0.819 \times 71.1 = 58.3\text{ g.}\)

Step 5: Calculate mass of \(\mathrm{O}_2\) produced

From the balanced equation, \(\frac{3}{2}\) or 1.5 moles of \(\mathrm{O}_2\) are produced for every 1 mole of \(\mathrm{CO}_2\). Therefore, \(0.409 \times 1.5 = 0.614\text{ moles of } \mathrm{O}_2\text{ are produced.}\)Oxygen's molar mass is 32 g/mol, so the mass of \(\mathrm{O}_2\) is \(0.614 \times 32 = 19.6\text{ g.}\)

Key Concepts

Oxidation Numbers

Understanding oxidation numbers is crucial in chemistry. These numbers tell us how many electrons an atom gains or loses during a chemical reaction. By assigning these numbers, we can determine which elements are oxidized and which are reduced.
In the reaction given, Potassium superoxide (\(\mathrm{KO}_{2}\)) has oxidation numbers of +1 for potassium (\(\mathrm{K}\)) and -1/2 for each oxygen atom. This unique fractional oxidation state arises because \(\mathrm{KO}_{2}\) contains a superoxide ion, where oxygen is partially reduced compared to most oxides. In carbon dioxide (\(\mathrm{CO}_{2}\)), carbon (\(\mathrm{C}\)) has an oxidation number of +4, while each oxygen (\(\mathrm{O}\)) is -2. Finally, in potassium carbonate (\(\mathrm{K}_{2}\mathrm{CO}_{3}\)), the cation \(\mathrm{K}\) stays at +1, carbon stays at +4, and oxygen remains at -2.
When examining these changes, we discover that there is no change in oxidation number for any element going from reactants to products. This means that no atoms are oxidized or reduced in our particular scenario. It is considered a non-redox reaction, even though oxygen gas (\(\mathrm{O}_{2}\)) is produced.

Stoichiometry

Stoichiometry allows us to calculate reactants and products in chemical reactions. By using ratios from a balanced chemical equation, we translate our reaction in terms of moles and grams.
In this exercise, we were provided with a balanced equation: \[4 \mathrm{KO}_{2}(s) + 2 \mathrm{CO}_{2}(g) \rightarrow 2 \mathrm{K}_{2}\mathrm{CO}_{3}(s) + 3 \mathrm{O}_{2}(g)\]This equation tells us two moles of \(\mathrm{KO}_{2}\) react with one mole of \(\mathrm{CO}_{2}\). It also tells us that for every two moles of \(\mathrm{CO}_{2}\), three moles of oxygen gas (\(\mathrm{O}_{2}\)) are produced.
Let's say you have 18 grams of \(\mathrm{CO}_{2}\). First, find the moles of \(\mathrm{CO}_{2}\) by dividing this mass by \(44 \text{ g/mol}\). With the moles of \(\mathrm{CO}_{2}\), you multiply by the stoichiometric ratio from the equation to find how many moles of \(\mathrm{KO}_{2}\) you need. Further, use its molar mass (\(71.1 \text{ g/mol}\)) to find the required grams of \(\mathrm{KO}_{2}\).
This method helps determine the exact amounts of reactants needed and predicts how much of a product, like \(\mathrm{O}_{2}\), forms.

Chemical Reactions

Chemical reactions involve the transformation of substances by breaking and forming chemical bonds. In these processes, atoms are rearranged, leading to the formation of new products.
To fully understand any chemical reaction, such as the reaction of potassium superoxide (\(\mathrm{KO}_{2}\)) with carbon dioxide (\(\mathrm{CO}_{2}\)), one must identify reactants and products and understand their functions.
In an oxygen mask scenario, \(\mathrm{KO}_{2}\) serves as a crucial reactant. It reacts with \(\mathrm{CO}_{2}\) to generate oxygen (\(\mathrm{O}_{2}\)), ensuring a breathable environment in closed or hazardous scenarios. The product \(\mathrm{K}_{2}\mathrm{CO}_{3}\) forms as part of this reaction, not as essential but as a consequence of the transformation.
Exploring these reactions highlights chemical relationships and allows us to harness them in practical applications, such as safety equipment for firefighters or astronauts.