4 KO2 + 2CO2 rightwards arrow 2 K2CO3 + 3 O2
How many moles of O2 can be produced from 12 moles of KO2 and 10 moles of CO2? If you were to produce 100.0 grams of O2 what would be the percent yield of the reaction?
This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants.
Using the coefficients in the balancaed equation, convert 12 mols KO2 to mols O2.
Do the same to convert mols CO2 to mols O2.
It is likely the two values will not be the same which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for the smaller number is the LR.
Using the smaller number, convert to grams O2. grams O2 = mols O2 x molar mass O2. This is the theoretical yield for O2.
Then percent yield = (g O2 produced/theoretical yield)*100 = ?
First find limiting reactant (KO2). The answer for the first question is 9 mols of oxygen.
For the second question 100/(9*16) * 100% = 69.4%
To determine the moles of O2 produced, we need to compare the stoichiometric ratio between KO2 and O2. According to the balanced equation:
4 KO2 + 2 CO2 → 2 K2CO3 + 3 O2
The ratio is 4:3, which means for every 4 moles of KO2 reacted, we will produce 3 moles of O2.
1) Moles of O2 from 12 moles of KO2:
Since the ratio is 4:3, we can set up a proportion to find the moles of O2 produced from 12 moles of KO2.
4 moles KO2 / 3 moles O2 = 12 moles KO2 / x moles O2
Solving for x gives us:
x = (12 moles KO2 * 3 moles O2) / 4 moles KO2
x = 9 moles O2
Therefore, 12 moles of KO2 can produce 9 moles of O2.
2) Moles of O2 from 10 moles of CO2:
Since the ratio between CO2 and O2 is 2:3, we can set up a similar proportion.
2 moles CO2 / 3 moles O2 = 10 moles CO2 / x moles O2
Solving for x gives us:
x = (10 moles CO2 * 3 moles O2) / 2 moles CO2
x = 15 moles O2
Therefore, 10 moles of CO2 can produce 15 moles of O2.
To calculate the percent yield, we need to compare the actual yield (moles or grams of O2 obtained) with the theoretical yield (moles or grams of O2 predicted from stoichiometry).
To find the theoretical yield in moles, we add the moles of O2 produced from the reactions with KO2 and CO2.
Theoretical yield = moles of O2 from KO2 + moles of O2 from CO2
Theoretical yield = 9 moles O2 + 15 moles O2
Theoretical yield = 24 moles O2
1 mole of O2 has a molar mass of 32 g, so the theoretical yield in grams is:
Theoretical yield (grams) = 24 moles O2 * 32 g/mol
Theoretical yield (grams) = 768 g O2
Given that the actual yield is 100.0 grams of O2, we can calculate the percent yield using the equation:
Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (100.0 g / 768 g) * 100
Percent yield = 13.02%
Therefore, the percent yield of the reaction is 13.02%.