Chemical Reaction Yield Calculation: C6H6 to C6H5NO2

What is the yield percentage of C6H5NO2 from the reaction?

A. 73.2%

Answer:

The yield percentage of the reaction of C6H6 with HNO3 to form C6H5NO2 is 73.2%

Yield percentage is a crucial factor in chemical reactions to determine the efficiency of the reaction process. In this case, we are analyzing the yield percentage of the reaction between C6H6 and HNO3 to produce C6H5NO2.

To calculate the yield percentage, we need to compare the actual yield (18.0g of C6H5NO2 obtained) with the theoretical yield. The theoretical yield is determined based on stoichiometry and the limiting reactant in the reaction.

First, we determine the molar masses of C6H6 and C6H5NO2. The molar mass of C6H6 is about 78 g/mol, while the molar mass of C6H5NO2 is about 123 g/mol.

Given that we have a 15.6g sample of C6H6, we can convert this mass to moles using the molar mass of C6H6. This gives us approximately 0.2 mol of C6H6. Since C6H6 is reacting with an excess of HNO3, it is the limiting reactant in this case.

Next, we calculate the theoretical yield of C6H5NO2 based on the moles of C6H6. Multiplying the moles of C6H6 by the molar mass of C6H5NO2 gives us a theoretical yield of approximately 24.6g.

Finally, using the formula for yield percentage: yield = (actual yield / theoretical yield) x 100%, we can plug in the values to find that the yield percentage of C6H5NO2 is 73.2%.