Answer:
There will be produced 0.8448 grams O2
Explanation:
<u>Step 1:</u> 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
<u>Step 2:</u> The balanced equation
4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)
<u>Step 3</u>: Calculate moles KO2
Moles KO2 = mass KO2 / Molar mass KO2
Moles KO2 = 2.50 grams / 71.1 g/mol
Moles KO2 = 0.0352 moles
<u>Step 4:</u> 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
<u>Step 5:</u> 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
<u>Step 6</u>: 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
<u>Step 7</u>: 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
