chemistry moles

Question

For the following reaction, 4.40 grams of potassium hydroxide are mixed with excess carbon dioxide. The reaction yields 4.04 grams of potassium carbonate.

carbon dioxide (g) + potassium hydroxide (aq) potassium carbonate (aq) + water (l)

What is the theoretical yield of potassium carbonate ?
What is the percent yield of potassium carbonate ?

Answer 1

1. First you need to write out the balanced chemical equation: 2KOH + CO₂ → K₂CO₃ + H₂O
2. The problem tells us we have excess CO₂ so that will definitely not be our limiting reagent and it means that if there is a limiting reagent it would have to be Potassium Hydroxide.
3. To find theoretical yield (the Maximum amount of product that can be formed) we must use KOH since it is potentially limiting
4. So first convert KOH (56.1 g/mol) to moles, (4.4 g KOH/ 56.1 g/mol)= 0.078 mol
5. Then use the stoichiometric coefficients to find the amount of moles of K₂CO₃ (138.2 g/mol) formed.
6. 0.078 moles KOH * (1 mol K₂CO₃ / 2 mol KOH)= .039 moles K₂CO_3.
7. Now convert to grams to find the theoretical yield.
8. .039 moles K₂CO_{3. *} (138.2 g/mol)= 5.41 g K₂CO_3.
9. 5.41 g is the theoretical yield for K₂CO_3. this reaction which means that if that reaction were to occur and not of the products were lost as heat or some other byproduct 5.41 g of K₂CO₃ would be produced_.
10. To find the percent yield you just do (actual yield/ theoretical yield) *100: (4.04g/ 5.41g) *100= 74.7%. This is the percent yield for K₂CO₃
11. Hopefully this is correct. feel free to reach out