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Gaseous ethane (CH3CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water . If 3.31g of carbon dioxide is produced from the reaction of 2.1g of ethane and 4.9g of oxygen gas, calculate the percent yield of carbon dioxide.

User Iqon
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Answer:

CO2 Percent Yield: 53.82%

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Gaseous ethane (CH3CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water . If 3.31g of carbon dioxide is produced from the reaction of 2.1g of ethane and 4.9g of oxygen gas, calculate the percent yield of carbon dioxide.

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To calculate the percent yield of carbon dioxide, we first need to determine the theoretical yield of carbon dioxide and then compare it to the actual yield. The theoretical yield is the maximum amount of product that can be formed based on the balanced chemical equation. The actual yield is the amount of product obtained in the experiment.

Let's start by writing the balanced chemical equation for the reaction:

C2H6 (g) + 7/2 O2 (g) → 2 CO2 (g) + 3 H2O (g)

Now, we need to calculate the molar masses of the substances involved:

Molar mass of C2H6 (ethane): 2 * (12.01 g/mol) + 6 * (1.01 g/mol) = 30.07 g/mol

Molar mass of O2 (oxygen gas): 2 * (16.00 g/mol) = 32.00 g/mol

Molar mass of CO2 (carbon dioxide): 12.01 g/mol + 2 * (16.00 g/mol) = 44.01 g/mol

Next, we'll calculate the number of moles for each reactant:

Moles of ethane (C2H6):

moles of C2H6 = mass of C2H6 / molar mass of C2H6

moles of C2H6 = 2.1 g / 30.07 g/mol ≈ 0.0699 mol

Moles of oxygen gas (O2):

moles of O2 = mass of O2 / molar mass of O2

moles of O2 = 4.9 g / 32.00 g/mol ≈ 0.1531 mol

Now, we need to determine the limiting reactant. The limiting reactant is the one that gets completely consumed in the reaction, thus limiting the amount of product that can be formed. To find the limiting reactant, we compare the mole ratios from the balanced equation.

From the balanced equation:

1 mol C2H6 requires 7/2 mol O2 to react.

Since we have:

Moles of C2H6 = 0.0699 mol

Moles of O2 = 0.1531 mol

The ratio of C2H6 to O2 is:

0.0699 mol / (0.1531 mol * 7/2) ≈ 0.508

The ratio for O2 to C2H6 is:

(0.1531 mol * 2 / 7) / 0.0699 mol ≈ 0.508

The ratio is approximately the same, indicating that the reaction is close to a 1:1 ratio between C2H6 and O2. Therefore, the limiting reactant is C2H6 (ethane).

Now, let's calculate the theoretical yield of CO2 based on the balanced equation:

1 mol C2H6 produces 2 mol CO2.

Moles of CO2 produced = 0.0699 mol of C2H6 * 2 mol CO2/1 mol C2H6 = 0.1398 mol CO2

Now, calculate the mass of the theoretical yield of CO2:

Mass of CO2 = moles of CO2 produced * molar mass of CO2

Mass of CO2 = 0.1398 mol * 44.01 g/mol ≈ 6.15 g

Finally, calculate the percent yield of carbon dioxide:

Percent yield = (Actual yield / Theoretical yield) * 100

Percent yield = (3.31 g / 6.15 g) * 100 ≈ 53.82%

So, the percent yield of carbon dioxide in this reaction is approximately 53.82%.

User Charma
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