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3 March, 17:59

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)

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  1. 3 March, 20:47
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    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
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