According to the following equation, Calculate the percentage yield if 550.0 g of toluene (C7H8 )added to an excess of nitric acid (HNO3) provides 305 g of the p-nitrotoluene (C7H7NO2 ) product in a lab experiment.
___C7H8 + ___HNO3  ___C7H7NO2 + ___H2O

37.69 % is the percentage yield if 550.0 g of toluene (C7H8 )added to an excess of nitric acid (HNO3) provides 305 g of the p-nitrotoluene (C7H7NO2 ) product in a lab experiment.

Explanation:

Balanced reaction equation:

C7H8 + ___HNO3  ___C7H7NO2 + ___H2O

data given:

mass of toluene C7H8 = 550 grams (limiting reagent)

mass of nitrotoluene obtained = 305 grams (actual yield)

From the reaction 1 mole of toluene react to form 1 mole of nitrotoluene

atomic mass of nitrotoluene is 137.14 grams/mole

mass of nitrotoluene produced is 137.14 grams

Now the number of moles of nitrotoluene taking part in the reaction is given by:

n =

  = 5.9  moles

hence, form the reaction when 1 mole toluene produced 1 mole of nitrotoluene

5.9 moles will produce 5.9 moles of nitrotoluene

This will give the yield according to the reaction i.e theoretical yield

mass of nitrotoluene = number of  moles x atomic mass of toluene

                                  = 5.9 x 137.14

                                   = 809.12 grams of nitrotoluene (theoretical yield)

Actual yield = 305 grams

formula for percent yield:

percent yield = x 100

                      = x 100

                      = 37.69 % is the yield percent.

37.26% is the correct answer.

Explanation: