If 40.0 grams of water are produced when 30.0 grams of ethane is burned, what is the % yield
To calculate the percent yield, you need to compare the actual yield of the reaction (in this case, the amount of water produced) to the theoretical yield (the maximum amount of water that should be produced based on stoichiometry).
First, you need to determine the balanced chemical equation for the reaction between ethane (C2H6) and oxygen (O2) to produce water (H2O) and carbon dioxide (CO2).
The balanced equation is:
C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O
According to the balanced equation, for every mole of ethane that reacts, you should get 3 moles of water. Also, the molar masses of ethane (C2H6) and water (H2O) are 30.07 g/mol and 18.02 g/mol, respectively.
Now, let's calculate the theoretical yield of water:
30.0 g ethane * (1 mol ethane / 30.07 g ethane) * (3 mol water / 1 mol ethane) * (18.02 g water / 1 mol water) = X g water
Simplifying this calculation, we get:
30.0 g ethane * (3 * 18.02 g water) / (30.07 g ethane) = X g water
X = 54.00 g water
The theoretical yield of water is 54.00 grams.
Now, let's calculate the percent yield:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Percent Yield = (40.0 g water / 54.00 g water) * 100 = 74.07%
Therefore, the percent yield is 74.07%.