Questions: Gaseous ethane (CH3CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 8.24 g of carbon dioxide is produced from the reaction of 5.41 g of ethane and 27.8 g of oxygen gas, calculate the percent yield of carbon dioxide. Round your answer to 3 significant figures.

Gaseous ethane (CH3CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 8.24 g of carbon dioxide is produced from the reaction of 5.41 g of ethane and 27.8 g of oxygen gas, calculate the percent yield of carbon dioxide.

Round your answer to 3 significant figures.

Transcript text: Gaseous ethane $\left(\mathrm{CH}_{3} \mathrm{CH}_{3}\right)$ reacts with gaseous oxygen gas $\left(\mathrm{O}_{2}\right)$ to produce gaseous carbon dioxide $\left(\mathrm{CO}_{2}\right)$ and gaseous water $\left(\mathrm{H}_{2} \mathrm{O}\right)$. If 8.24 g of carbon dioxide is produced from the reaction of 5.41 g of ethane and 27.8 g of oxygen gas, calculate the percent yield of carbon dioxide. Round your answer to 3 significant figures. (1\% $\square$ $\times$ o IDon't Know Submit

Solution

Solution Steps

Step 1: Balanced Chemical Equation

The balanced chemical equation for the combustion of ethane is:

2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g)

Step 2: Determine the Limiting Reactant

First, convert the given masses of ethane (5.41 g) and oxygen (27.8 g) to moles using their respective molar masses (C₂H₆ = 30.07 g/mol, O₂ = 32.00 g/mol):

Moles of C₂H₆ = 5.41 g / 30.07 g/mol = 0.180 mol

Moles of O₂ = 27.8 g / 32.00 g/mol = 0.869 mol

Next, divide the moles of each reactant by their respective stoichiometric coefficients:

C₂H₆: 0.180 mol / 2 = 0.0900

O₂: 0.869 mol / 7 = 0.124

Since the value for ethane (0.0900) is smaller, ethane is the limiting reactant.

Step 3: Theoretical Yield of CO₂

Using the stoichiometry of the balanced equation, calculate the theoretical yield of CO₂ from the moles of limiting reactant (ethane):

Moles of CO₂ = 0.180 mol C₂H₆ * (4 mol CO₂ / 2 mol C₂H₆) = 0.360 mol CO₂

Convert moles of CO₂ to grams using the molar mass of CO₂ (44.01 g/mol):

Theoretical yield of CO₂ = 0.360 mol * 44.01 g/mol = 15.8 g

Step 4: Calculate Percent Yield