Question $\underline{\text { Open Ebook section } 7.4}$
Potassium superoxide, $\mathrm{KO}_{2}$ , reacts with carbon dioxide to form potassium carbonate and oxygen: $4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{~K}_{2} \mathrm{CO}_{3}+3 \mathrm{O}_{2}$
3rd attempt See Periodic Table
This reaction makes potassium superoxide useful in a self-contained breathing apparatus. How much $\mathrm{O}_{2}$ could be produced from 2.59 g of $\mathrm{KO}_{2}$ and 4.60 g of $\mathrm{CO}_{2}$ ?

Studdy Solution

STEP 1

1. The chemical reaction is balanced as given: $4 \text{ KO}_2 + 2 \text{ CO}_2 \rightarrow 2 \text{ K}_2\text{CO}_3 + 3 \text{ O}_2$ .
2. We are given 2.59 g of $\text{KO}_2$ and 4.60 g of $\text{CO}_2$ .
3. We need to determine the amount of $\text{O}_2$ produced.
4. Molar masses are required: $\text{KO}_2$ and $\text{CO}_2$ .

STEP 2

1. Calculate the molar masses of $\text{KO}_2$ and $\text{CO}_2$ .
2. Convert the given masses of $\text{KO}_2$ and $\text{CO}_2$ to moles.
3. Determine the limiting reactant.
4. Calculate the moles of $\text{O}_2$ produced based on the limiting reactant.
5. Convert moles of $\text{O}_2$ to grams.

STEP 3

Calculate the molar masses of $\text{KO}_2$ and $\text{CO}_2$ .
- Molar mass of $\text{KO}_2$ : $K = 39.10 \, \text{g/mol}, O = 16.00 \, \text{g/mol}$ $ \text{Molar mass of } \text{KO}_2 = 39.10 + 2 \times 16.00 = 71.10 \, \text{g/mol} \]
- Molar mass of $\text{CO}_2$ : $C = 12.01 \, \text{g/mol}, O = 16.00 \, \text{g/mol}$ $ \text{Molar mass of } \text{CO}_2 = 12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol} \]

STEP 4

Convert the given masses to moles.
- Moles of $\text{KO}_2$ : $ \text{Moles of } \text{KO}_2 = \frac{2.59 \, \text{g}}{71.10 \, \text{g/mol}} \approx 0.0364 \, \text{mol} \]
- Moles of $\text{CO}_2$ : $ \text{Moles of } \text{CO}_2 = \frac{4.60 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.1045 \, \text{mol} \]

STEP 5

Determine the limiting reactant.
From the balanced equation: $4 \text{ KO}_2 + 2 \text{ CO}_2 \rightarrow 2 \text{ K}_2\text{CO}_3 + 3 \text{ O}_2$
- Ratio of $\text{KO}_2$ to $\text{CO}_2$ is 4:2 or 2:1. - Calculate the required moles of $\text{CO}_2$ for 0.0364 moles of $\text{KO}_2$ : $ \text{Required moles of } \text{CO}_2 = \frac{0.0364}{2} = 0.0182 \, \text{mol} \]
Since $0.1045 \, \text{mol}$ of $\text{CO}_2$ is available, $\text{KO}_2$ is the limiting reactant.

STEP 6

Calculate the moles of $\text{O}_2$ produced based on the limiting reactant.
From the balanced equation, 4 moles of $\text{KO}_2$ produce 3 moles of $\text{O}_2$ .
- Moles of $\text{O}_2$ produced: $ \text{Moles of } \text{O}_2 = \frac{3}{4} \times 0.0364 \approx 0.0273 \, \text{mol} \]

STEP 7

Convert moles of $\text{O}_2$ to grams.
- Molar mass of $\text{O}_2$ : $2 \times 16.00 = 32.00 \, \text{g/mol}$
- Grams of $\text{O}_2$ : $ \text{Grams of } \text{O}_2 = 0.0273 \, \text{mol} \times 32.00 \, \text{g/mol} \approx 0.8736 \, \text{g} \]
The amount of $\text{O}_2$ produced is approximately $0.874 \, \text{g}$ .

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