Question

4KO2(s) +2CO2(g) (arrow) 2k2CO3(s)+ 3O2

suppose (9.1250x10^1) grams of KO2 were added to (5.52x 10^1) grams of CO2 how much in g of k2CO3 was produced once the reaction was run to completion?.

Answer 1

1. Calculate the moles of KO2 and CO2:
Moles of KO2 = (mass of KO2) / (molar mass of KO2)
Moles of KO2 = (9.1250 x 10^1 g) / (71.10 g/mol) ≈ 1.282 moles
Moles of CO2 = (mass of CO2) / (molar mass of CO2)
Moles of CO2 = (5.52 x 10^1 g) / (44.01 g/mol) ≈ 1.255 moles

2. Determine the limiting reactant:
To find the limiting reactant, compare the mole ratio of KO2 to CO2 from the balanced equation. The balanced equation shows that 4 moles of KO2 react with 2 moles of CO2.
(Moles of KO2) / 4 = (Moles of CO2) / 2
1.282 / 4 = 1.255 / 2
Since the ratio on the left is smaller, KO2 is the limiting reactant.

3. Use the stoichiometry of the balanced equation to find moles of K2CO3:
From the balanced equation, 4 moles of KO2 produce 2 moles of K2CO3. So, for 1.282 moles of KO2, we get:
Moles of K2CO3 = (1.282 moles KO2) x (2 moles K2CO3 / 4 moles KO2) = 0.641 moles K2CO3

4. Convert moles of K2CO3 to grams:
Moles of K2CO3 = 0.641 moles
Molar mass of K2CO3 = 138.21 g/mol (from the molar masses of K, C, and O)
Mass of K2CO3 = (0.641 moles) x (138.21 g/mol) ≈ 88.54 grams

So, approximately 88.54 grams of K2CO3 would be produced when the reaction is run to completion.