Gaseous ethane (CH3CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 2.71 g of carbon dioxide is produced from the reaction of 3.31 g of ethane and 23.9 g of oxygen gas, calculate the percent yield of carbon dioxide.

The percent yield is 28.0 %

Explanation:

Step 1: Data given

Mass of carbon dioxide = 2.71 grams

Mass of ethane = 3.31 grams

Mass of oxygen = 23.9 grams

Molar mass ethane = 30.07 g/mol

Molar mass CO2 = 44.01 g/mol

Molar mass O2 = 32.0 g/mol

Step 2: The balanced equation

2C2H6 + 7O2 → 4CO2 + 6H2O

Step 3: Calculate moles

Moles = mass /molar mass

Moles ethane = 3.31 grams / 30.07 g/mol

Moles ethane = 0.110 moles

Moles oxygen = 23.9 grams / 32.0 g/mol

Moles oxygen = 0.747 moles

Step 4: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

Ethane is the limiting reactant. It will completely be consumed (0.110 moles)

Oxygen is the limiting reactant. There will react 0.110 *3.5 = 0.385 moles

There will remain 0.747 - 0.385 = 0.362 moles O2

Step 5: Calculate moles CO2

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

For 0.110 moles ethane we'll have 2*0.110 = 0.220 moles CO2

Step 6: Calculate theoretical yield of CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.220 moles * 44.01 g/mol

Mass CO2 = 9.68 grams

Step 7: Calculate % yield

% yield = (actual yield / theoretical yield) * 100 %

% yield =(2.71 grams / 9.68 grams ) * 100 %

% yield =  28.0 %

The percent yield is 28.0 %

c) should be the answer.