Theoretical yield

Question

What is the theoretical yield of H2O (MM=18) produces by burning 150g of C2H6 (MM=30) and enough O2 to allow for complete reaction. If the actual yield of H2O was found to be 135g of H2O, what is the % yield of H2O?

Answer 1

Hi, Allison,

Burning ethane again, are we?? 😊 O.K., so here’s what we need to do:

(1)   Write a balanced equation for this combustion reaction like so:

2C₂H₆ + 7O₂ -> 4CO₂ + 6H₂0

(2)   Convert the 150g of C₂H₆ (aka ethane, you know) into a molar quantity:

150g of ethane / 30 g/mol ethane = 5.0 mole C₂H₆

(3)   Solve the first equation to calculate theoretical yield of water produced:

If 2 mole C₂H₆ would generate 6 moles water, then

1 mole C₂H₆ would generate 3 moles of water, and

5 moles C₂H₆ would generate 15 moles of water.

15 moles of water weighs: (15 moles) x (18 g/mole) or 270 grams. This is the theoretical yield.

(4)   If the actual yield of H₂O was found to be 135g of H₂O, what is the % yield of H₂O?

I’ll bet you saw this coming, eh? The percentage yield would be 135g/270g or 50%.

Done!

Chris

Answer 2

Write the balanced equation for compete combustion of C2H6:

2C2H6 + 7O2 ==> 4CO2 + 6H2O ... balanced equation

Use dimensional analysis to find theoretical yield of H2O:

150 g C2H6 x 1 mol/30 g x 6 mol H2O/2 mol C2H6 x 18 g H2O/mol = 270 g H2O

Percent yield = actual yield/theoretical yield (x100%):

135 g/270 g (x100%) = 50% yield