Question

The directions tell you to add 102 mg of methyl benzoate (mw =136 g/mol) to a vial. seperately, .50 mL of conc. sulfuric acid and 0.80 mL of conc. nitric acid are put in a vial to form the nitrating mixture. which is the limiting reagent? show your work.

Answer 1

We use concentration of nitric and sulfuric acid

[HNO3] = 15.8 M

[H2SO4]= 18.4 M

Mass of methyl benzoate = 102 mg

Solution:

Lets show this reaction :

This is nitration reaction. Lets show how NO2 group is formed by the reaction of H2SO4 and HNO3.

H2SO4 + HNO3 --- > NO₂⁺ + H3O+ + HSO4

So the mole ration of all the reactants = 1: 1 : 1

Now we have to find the least number of moles among these three.

Calculation of moles:

Moles of methyl benzoate:

Moles = mass in g /molar mass

= 0.102 g x 1/ molar mass of methyl benzoate

= 0.102 g x 1 mol / 136 g

= 0.00075 mol

Moles of H2SO4

Moles   = volume in L x molarity

Moles of H2SO4 = 0.00050 L x 18.4 M = 0.0092 mol H2SO4

Moles of Nitric acid = 0.00080 L x 15.8 M = 0.01264 mol

From the calculation we can see that methyl benzoate is limiting reactant since it has least number of moles.