Question

Potassium superoxide, KO2, reacts with carbon dioxide to form potassium carbonate and oxygen: This reaction makes potassium superoxide useful in a self-contained breathing apparatus. How much O2 could be produced from 2.50 g of KO2 and 4.50 g of CO2?4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)

Answer 1

Answer:

There will be produced 0.8448 grams O2

Explanation:

Step 1: Data given

Mass of KO2 = 2.50 grams

Mass of CO2 = 4.50 grams

Molar mass KO2 = 71.1 g/mol

Molar mass of CO2 = 44.01 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)

Step 3: Calculate moles KO2

Moles KO2 = mass KO2 / Molar mass KO2

Moles KO2 = 2.50 grams / 71.1 g/mol

Moles KO2 = 0.0352 moles

Step 4: Calculate moles of CO2:

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 4.50 grams / 44.01 g/mol

Moles CO2 = 0.102 moles

Step 5: calculate limiting reactant

For 4 moles of KO2 consumed, we need 2 moles of CO2 to produce 2 moles of K2CO3 and 3 moles of O2

KO2 is the limiting reactant. It will completely be consumed. (0.0352 moles).

CO2 is in excess. There will react 0.0352/2 = 0.0176 moles of CO2

There will remain 0.102 - 0.0176 = 0.0844 moles CO2

Step 6: Calculate moles O2 produced

For 4 moles of KO2 consumed, we need 2 moles of CO2 to produce 2 moles of K2CO3 and 3 moles of O2

For 0.0352 moles KO2 consumed, we have 3/4 * 0.0352 = 0.0264 moles of O2 produced

Step 7: Calculate mass of 02 produced

Mass O2 produced = Moles O2 * Molar mass 02

Mass O2 produced = 0.0264 moles * 32 g/mol

Mass O2 produced = 0.8448 grams O2

There will be produced 0.8448 grams O2