Nitrotoluene Yield Question

[orgostudent]

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?

[Phi for All]

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[Phi for All]

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Moderator Note

Please see this thread on the same question.