For the following reaction, 4.71 grams of oxygen gas are mixed with excess glucose. The reaction yields 5.53 grams of carbon dioxide.

Question

For the following reaction, 4.71 grams of oxygen gas are mixed with excess glucose. The reaction yields 5.53 grams of carbon dioxide.

glucose (s) + oxygen (g) carbon dioxide (g) + water (l)

What is the theoretical yield of carbon dioxide ? ____ grams

What is the percent yield of carbon dioxide ? __ %

Answer 1

To answer this question, you must first balance the equation for the reaction.

Glucose + oxygen creates carbon dioxide + water

C₆H₁₂O₆ + O₂ =  CO₂ + H₂O

Note that the left side contains 6 C atoms, 12 H atoms and a total of 8 O atoms and the right side contains 1 C atom, 3 O atoms, and 2 H atoms.

With this type of equation, you typically balance the C atoms first, then balance the H atoms, and lastly balance the O atoms.

The balanced equation is determined by adding a coefficient of 6 to the CO₂ to balance the carbon atoms, then adding a coefficient of 6 to H₂O to balance the hydrogen atoms.

The right side now has a total of 18 O atoms and the left side has 8 O atoms. You need to add 10 more O atoms to the left side. You do this by giving O₂ a coefficient of 6.

The balanced equation looks like this:

C₆H₁₂O₆ + 6O₂ = 6CO₂ + 6H₂O

Now you are prepared to answer the questions.

1. To determine the theoretical yield of CO₂, calculate the number of moles of O₂, then multiply this by the ratio of moles of O₂ : moles of CO₂ from their coefficients in the balanced equation, then multiply this by the molar mass of CO₂.

(4.71 g of O₂) divided by (32.0 g/mole of O₂) = 0.147 moles of O₂

(0.147 moles of O₂) x (6 moles of O₂ / 6 moles of CO₂) x (44.01 g/mole of CO₂) = 6.47 g of CO₂ as the theoretical yield of CO₂.

2.The % yield of CO₂ = (Actual yield of CO₂ divided by Theoretical yield of CO₂) x 100%).

= (5.53 g / 6.47g) x 100% = 85.5%