I am trying to find the theoretical yield and percent yield in nitrotoluene.

 

The question: Exactly 4.0 mL of toluene (density = 0.87 g/mL) were reacted with excess nitric acid in the presence of excess sulfuric acid, and 3.6 g of nitrotoluene were obtained.

 

So I started with the balanced equation: C₇H₈ + HNO₃ --> C₇H₇NO₂ + H₂O

 

Then I calculated the grams of toluene: 4.0mL * 0.87g/mL = 3.48g C₇H₈

 

Then I calculated the moles of toluene: 3.48g * 1 mole/92.138g = 0.038 mole C₇H₈

 

Then I calculated the moles of nitrotoluene: 3.6g * 1 mole/137.136g = 0.026 mole C₇H₇NO₂

 

This equation is a 1:1 ratio so I can't multiply by anything, so trying to find the limiting reactant is throwing me off especially since I don't know how much nitric acid is being used. Any suggestions on how to proceed?

Edited February 24, 201312 yr by cdornz

The problem specified that nitric acid and sulfuric acid were in excess (sulfuric was probably a catalyst, anyway). Therefore they cannot be limiting reagents by the assumptions of the problem.

According to the equation 0.038 mol toluene should produce 0.038 mol nitrotoluene.

You obtained 0.026mol nitrotoluene, so what is the actual yield in % and what is the theoretical yield in grams of nitrotoluene?