Question

Limiting reagent and theoretical yield, product: m-nitrobenzate

Reaction : 1.5 grams m-nitrobenzoic acid, 12 mL methanol, .6 mL sulfuric acid with conc. of 18.4 mol/L

I started the calcualtions but I wasnt sure where to go next:

.0006 L H2SO4 × 18.4 mol/L concentration H2SO4 = .01104 mol H2SO4

Answer 1

Product is Methyl 3-nitrobenzoate with MW = 181.145 g/mol

MW of m-nitrobenzoic acid = 167.12 g/mol

1) moles of m-nitrobenzoic acid = g/MW = 1.5/167.12 = 0.00898 moles

2) 6 mL H2SO4 = 0.006 L H2SO4 × 18.4 mol/L concentration H2SO4 = 0.1104 mol H2SO4

3) Density of methanol = 0.792 g/mL

Gms of methanol = Density X Volume = 0.792 X 12 = 9.504 gms methanol

moles of methanol = 9.504/32.04 = 0.2966 moles.

From above calculation, the moles of m-nitrobenzoic acid is lowest. Hence it is the limiting agent.

D) Theoratical yield

167.12 gms of m-nitrobenzoic acid gives  181.145 gms of methyl-3-nitrobenzoate

then, 1.5 gms of m-nitrobenzoic acid will give

= 1.5 X 181.145/167.12 = 1.62 gms of methyl-3-nitrobenzoate

Theoratical yield = 1.62 gms