The combustion of ethane produces carbon dioxide and water. If 27.6 g C2H6 are burned in the presence of excess air (O2) and 76.7 g of carbon dioxide are produced, what is the percent yield?
2 C2H6 + 7 02 --> 4 CO2 + 6 H20
divide by molar mass to find moles of ethane
... 27.6 / molar mass
find expected moles of CO2 ... twice ethane
... use molar mass to find expected yield
% yield = 76.7 / expected yield
To find the percent yield of a reaction, we need to compare the actual yield (the amount of product obtained in the experiment) to the theoretical yield (the amount of product that should be obtained according to stoichiometry calculations).
In this case, we are given that 27.6 g of ethane (C2H6) are burned, and 76.7 g of carbon dioxide (CO2) are produced. We need to calculate the theoretical yield of carbon dioxide based on the number of moles of ethane burned.
To find the number of moles of ethane, we can use its molar mass. The molar mass of C2H6 is calculated as follows:
C: 2 atoms x atomic mass of C = 2 x 12.01 g/mol = 24.02 g/mol
H: 6 atoms x atomic mass of H = 6 x 1.01 g/mol = 6.06 g/mol
Total molar mass of C2H6 = 24.02 g/mol + 6.06 g/mol = 30.08 g/mol
Now, we can calculate the number of moles of ethane:
Number of moles of ethane = mass of ethane / molar mass of ethane
= 27.6 g / 30.08 g/mol
= 0.918 mol
According to the balanced chemical equation for the combustion of ethane, the ratio of moles of ethane to moles of carbon dioxide is 1:2. Therefore, the theoretical yield of carbon dioxide can be calculated as follows:
Number of moles of carbon dioxide = 2 x Number of moles of ethane
= 2 x 0.918 mol
= 1.836 mol
Now, we can calculate the theoretical yield of carbon dioxide in grams:
Theoretical yield of carbon dioxide = number of moles of carbon dioxide x molar mass of carbon dioxide
= 1.836 mol x 44.01 g/mol
= 80.74 g
Finally, we can calculate the percent yield:
Percent yield = (actual yield / theoretical yield) x 100
= (76.7 g / 80.74 g) x 100
= 95.0%
Therefore, the percent yield of the reaction is 95.0%.