Volume of gas question

Question

The combustion of propane in oxygen is described by the following equation: C3H8 + 5O2 -> 3CO2 + 4H2O

If the reaction of 5g propane proceeds in 75% yield

b. What volume of CO2 is produced?

I worked out the moles of propane, multiplied them by 3 and then multiplied that by 24 and then divided that by 100 and multiplied by 75 to get 75% but that's not correct. Really confused on this question, any help is appreciated.

The answer is 5.72L

Answer 1

(There doesn't seem to be a part a. to this question)

The theoretical yield for moles of CO2 will be 3 times moles of propane. Correct.

So 3 X 5/44.0 = 0.341 moles CO2. However, the actual yield is 75% of that, 0.256 moles.

The problem doesn't state the temperature or pressure, so we will guess that this is at STP conditions, in which case the molar volume of an ideal gas is about 22.4 L. Multiply this by 0.256 and get 5.73 L. There could be rounding errors.

Also, it should be noted that the data as given has the yield with 2 sig figs, and the mass of propane is only given to 1 sig fig. So, unless the data was actually given more precisely, such as 75.0 and 5.00, writing the answer with 3 sig figs is not justified. Really, you would be limited to 1 sig fig, which would require rounding to 6 L.