Percent Yield of Carbon Dioxide?

Question

Liquid octane CH3CH26CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2

and gaseous water H2O. If 46.1g of carbon dioxide is produced from the reaction of 63.97g of octane and 145.6g of oxygen gas, calculate the percent yield of carbon dioxide.

Be sure your answer has the correct number of significant digits in it.

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I originally tried to do this question alone but I'm continuously getting the incorrect answer. I'm not sure what I'm doing wrong, it would be helpful to see the process.

Answer 1

It would be helpful to show us what you did so that we could explain where you might have gone wrong. Maybe next time, eh?

Write a correctly balanced equation:

CH3(CH2)6CH3 + O2 ==> CO2 + H2O can be written as...

2C₈H₁₈ + 25O₂ ==> 16CO₂ + 18H₂O ... balanced equation for combustion of octane

Find limiting reactant, if any.

moles C₂H₁₈ = 63.97 g x 1 mol/114.2 g = 0.5602 moles C₈H₁₈ (÷ 2 -> 0.2801)

moles O₂ = 145.6 g O₂ x 1 mol/32 g = 4.55 moles O₂ (÷ 25 -> 0.182)

O₂ is LIMITING

Theoretical yield is based on the limiting reactant, and as such, we have...

4.55 moles O₂ x 16 moles CO₂ / 25 moles O₂ x 44 g CO₂/mole CO₂ = 128 g CO₂ as theoretical yield

Percent yield = actual yield/theoretical yield (x100) = 46.1 g / 128 g (x100) = 36.0% yield