Question

Can anyone help me with this problem?
Calculate the mass of CO2 that can be produced if the reaction of 46.9 g of pentane and sufficient oxygen has a 61.0 % yield.

[Pentane (C5H12) burns in oxygen to produce carbon dioxide and water.
C5H12(g)+8O2(g)→5CO2(g)+6H2O(g)]

Answer 1

First we have to work out the moles of CO2. We can do this by first working out the moles of pentane using this equation: moles = mass/molar mass

46.9g/72g mol-1 = 0.651mol (3 sig fig) of C5H12

Now the answer we have worked out is the moles for 1mol of C5H12. We need to use this and deduce the moles to get the answer for the moles of CO2. The answer we have is for 1 mole, however CO2 is 5 mol. This means we will have to times our answer by 5.

0.651mol x 5 = 3.26mol (3 sig fig) of CO2 at 100% percentage yield.

Now we will work out the mass of CO2 at 100% yield by using this equation: mass = moles x molar mass

3.26mol x 44g mol-1 = 143.44g at 100%

Now the answer we have is for 100% percentage yield, however we need it for 61.0% yield. We can work out the mass for 61% yield by using this equation: mass at 100% x yield needed = mass at yield needed

143.44g x 0.61 = 87.5g (3 sig fig)

So the answer us 87.5g at 61%. I hope this is right and I hope it helped!

Answer 2

Wow, thank you for all the descriptive steps! This helped so much :3

Answer 3

No problem! Glad I could help :)