With 143.6 grams of acetylene and an excess amount of oxygen gas, what is the percent yeild of carbon dioxide if the actual yield was 452.78 grams? Balanced equation 2C2H2 + 5O2 --> 4CO2 + 2H20

93.28%

Explanation:

To solve the percent yield we need to find theoretical yield:

Percent yield = Actual Yield (452.78g) / Theoretical yield * 100

Theoretical yield is obtained converting the mass of acetylene to moles and using the balanced equation determine the moles of CO₂ produced assuming a 100% of reaction:

Moles acetylene (Molar mass: 26.04g/mol)

143.6g C₂H₂ * (1mol / 26.04g) = 5.515 moles C₂H₂

Moles CO₂:

5.515 moles C₂H₂ * (4 moles CO₂ / 2mol C₂H₂) = 11.029moles CO₂

Mass CO₂ (Molar mass: 44.01g/mol):

11.029moles CO₂  * (44.01g / mol) = 485.39g CO₂ is theoretical yield

Percent yield is:

Percent yield = 452.78g / 485.39g * 100

= 93.28%

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