When 16 g of methane (CH4) and 32 g of oxygen (O2) reacted to produce carbon dioxide and water, 11 g of carbon dioxide was produced. Calculate the percent yield of carbon dioxide in this reaction.

Question

Chemistry

Jayden Walton

20 March, 20:27

When 16 g of methane (CH4) and 32 g of oxygen (O2) reacted to produce carbon dioxide and water, 11 g of carbon dioxide was produced. Calculate the percent yield of carbon dioxide in this reaction.

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Answers (1)

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Cohen · Answer 1

The yield is 50%

Explanation:

The equation for this reaction is as follows:

CH₄ + O₂ → CO₂ + H₂O

Let's balance the equation:

We have 1C on each side of the equation, so the C is balanced.

We have 4 H on the left side of the equation and 2 on the right side. Then, a 2 should be added to H₂O (2 (H₂O)). Now we have four on each side and the H is balanced.

Finally, we have 2O on the left side and 4 on the right side (2 from CO₂ and 2 from 2 (H₂O)), then a 2 should be added to O₂ in the left side (2O₂).

This is the balanced equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

The mass of a mol (molar mass) of methane is 16 g (this is the mass of 6.022 x 10²³ molecules of methane). It is calculated adding the mass of a mol of C (12 g) plus the mass of 4 mol of H (4 g). Putting it simply, the molar mass can be calculated by adding the atomic weight of each atom that composes the molecule in grams. In the case of methane (C = 12, 4H = 4*1, molar mass CH₄ = 16 g)

In the same way, 1 mol of O₂ has a mass of 32 g (16 (the atomic weight of oxygen) x 2).

In our problem, we have that 16 g methane (1 mol) reacts with 32 g O₂ (also 1 mol O₂ (remember 16 x 2)). But from our balanced equation, we know that 1 mol methane needs 2 mols O₂ to produce a complete reaction. But we have only half that quantity of oxygen. It means that only 0.5 mol methane will react. The amount of O₂ is not sufficient to react with all the methane. This lack of oxygen is limiting the reaction from completion and, therefore, oxygen is the limiting reagent of the reaction.

All subsequent calculations should be done using the limiting reagent, O₂. This is the reagent that reacts completely.

From the equation, we know that 2 mol O₂ produce 1 mol CO₂ when oxygen reacts with methane. Since we only have 1 mol oxygen, only 0.5 mol CO₂ would be produced, if the reaction has a yield of 100%.

Molar mass CO₂ = 12 + 16*2 = 44 g

mass of 0.5 mol CO₂ = 44 g / 2 = 22 g

In our problem we obtained 11 g carbon dioxide, so if 22 g correspond to a yield of 100%, 11 g would be a yield of (11 g * 100% / 22 g) 50%.