determine the percent yield when 5.00g benzene burns in oxygen gas to form 1.50g of carbon dioxide and water vapor.
2C6H6 + 15O2 ==> 12CO2 + 6H2O
mols benzene = grams/molar mass
Using the coefficients in the balanced equation, convert mols benzene to mols CO2.
Convert mols H2O to grams H2O with g = mols x molar mass. This is the theoretical yield (TY). The actual yield (AY) in the problem is 1.50g.
%yield = (AY/TY)*100 = ?
To determine the percent yield in a chemical 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 have been obtained if the reaction went to completion).
In this case, we are given that 5.00g of benzene (C6H6) reacts with oxygen gas (O2) to form 1.50g of carbon dioxide (CO2) and water vapor (H2O).
First, we need to write the balanced chemical equation for the reaction:
C6H6 + O2 -> CO2 + H2O
From the balanced equation, we can see that the ratio of benzene to carbon dioxide is 1:1.
Next, we calculate the theoretical yield of carbon dioxide using the molar mass of benzene and the molar mass of carbon dioxide.
1 mol of benzene (C6H6) has a molar mass of 78.11 g/mol.
1 mol of carbon dioxide (CO2) has a molar mass of 44.01 g/mol.
Since the ratio of benzene to carbon dioxide is 1:1, we can directly convert the mass of benzene to moles of carbon dioxide:
5.00 g of benzene * (1 mol benzene / 78.11 g benzene) * (1 mol CO2 / 1 mol benzene) * (44.01 g CO2 / 1 mol CO2) = theoretical yield of carbon dioxide in grams
Calculating this, we find that the theoretical yield of carbon dioxide is 3.57 g.
Lastly, we calculate the percent yield using the following formula:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
In this case, the actual yield is given as 1.50 g, and the theoretical yield is 3.57 g.
Percent Yield = (1.50 g / 3.57 g) * 100 = 42% (rounded to two decimal places)
Therefore, the percent yield of the reaction is approximately 42%.