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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2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.