For the following reaction, 17.1 grams of carbon dioxide are allowed to react with 47.4 grams of potassium hydroxide .

Question

For the following reaction, 17.1 grams of carbon dioxide are allowed to react with 47.4 grams of potassium hydroxide .

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

What is the maximum amount of potassium carbonate that can be formed? _____ grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? _____ grams

Answer 1

Carbon dioxide + potassium hydroxide reacts to form potassium + water

The unbalanced equation is:

CO₂ + KOH = K₂CO₃ + H₂O

This equation can be balance by giving KOH a coefficient of 2.

CO₂ + 2KOH = K₂CO₃ + H₂O

To determine the maximum amount of K₂CO₃ formed, you must first determine the number of moles of each reagent you have and then use the coefficients from the balanced equation to determine which is present in excess. The amount of limiting reagent will determine the amount of K₂CO₃ product produced.

Moles = mass in grams divided by molar mas

Moles of CO₂ present = (17.1 g of CO₂) divided by (44.01 g/ moles of CO₂) = 0.389 moles of CO₂

Moles of KOH present = (47.4 g of CO₂) divided by (56.11 g/mole of KOH) = 0.845 moles of KOH

KOH reacts with CO₂ in a 2:1 ratio,

Moles of KOH needed to react with 0.389 moles of CO₂ = 0.389 moles of CO₂ x ( 2 moles of KOH divided by 1 mole of CO₂)= 0.389 moles x (2) = 0.778 moles of KOH

You have more than enough KOH for the amount of CO₂, so the amount of CO₂ will determine the amount of K₂CO₃ produced.

Since the balance reaction shows a 1:1 ratio of CO₂ : K₂CO₃, the 0.389 moles of CO₂ present will produce 0.389 moles of K₂CO₃.

0.389 moles of K₂CO₃ x molar mass of K₂CO₃ gives the # grams of K₂CO₃ that can be produced.

0.389 moles x 138.03 g/mole = 53.7 g of K₂CO₃ that can be formed.

As explained above in bold, CO₂ (carbon dioxide) is the limiting reagent.

KOH is the excess reagent: (Starting Moles of excess reagent - moles of excess reagent needed for the reaction ) x molar mass of the excess reagent gives the amount of excess reagent in grams that is present when the reaction is complete.

(0.845 moles of KOH - 0.778 moles KOH) x 56.11 g/mole of KOH = 0.067 moles x 56.11 g/mole = 3.76 g KOH left after completion of the reaction.