A student extracts 6.3 g of water when performing the following combustion:
C2H8 + 4O2 --> 2 CO2 + 4H2O
If the theoretical yield is 8.6, what is the percent yield?
%yield = (actual yield/theor yield)*100 = ?
To calculate the percent yield, you need to compare the actual yield of a reaction to the theoretical yield. The theoretical yield is the maximum amount of product that can be obtained from a given amount of reactant, assuming that the reaction goes to completion. The actual yield is the amount of product obtained in an experiment.
In this case, the student extracted 6.3 g of water. To find the percent yield, we need to compare this to the theoretical yield of water.
From the balanced equation, we can see that the molar ratio between C2H8 (ethane) and H2O (water) is 1:4. This means that for every 1 mole of ethane, 4 moles of water are produced.
To calculate the theoretical yield of water, we can use the molar mass of ethane (C2H8) and the molar mass of water (H2O).
The molar mass of C2H8 is calculated as:
(2 x 12.01 g/mol) + (8 x 1.01 g/mol) = 30.07 g/mol
The molar mass of H2O is simply the sum of the atomic masses of hydrogen (1.01 g/mol) and oxygen (16.00 g/mol), which gives:
(2 x 1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Now we can calculate the theoretical yield of water. We'll use the following equation:
(6.3 g C2H8) x (4 mol H2O/1 mol C2H8) x (18.02 g H2O/1 mol H2O) = X g H2O
Simplifying this calculation, we find:
X = (6.3 g C2H8) x (4 mol H2O/1 mol C2H8) x (18.02 g H2O/1 mol H2O) = 453.6 g H2O
Therefore, the theoretical yield of water is 453.6 g.
To calculate the percent yield, we can use the following equation:
Percent Yield = (Actual Yield/Theoretical Yield) x 100%.
Substituting the given values:
Percent Yield = (6.3 g/453.6 g) x 100% = 1.39%
So, the percent yield is 1.39%.