Theoretical Yield of Nitrotoluene

[cdornz]

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, 2013 by cdornz

[BabcockHall]

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.

[Knumbnuts]

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?