Question

assuming that in the hydrolysis step you have 1 mole of p-nitroacetanilide and 300 mL of 6 M HCl. After the neutralization, you end up with 100 g of p-nitroaniline. a)what is the limiting reactant? B)percent yeild

Answer 1

p-nitroacetanilide + 6 M HCl ----> p-nitroaniline

no of moles of p-nitroacetanilide = 1 moles

no of moles of HCl = molarity of HCl x volume of HCl in liters = 6 M x 0.3L = 1.8 mole

for one mole of p-nitroacetanilide 1.8 mole of HCl taken means limiting agent is p-nitroacetanilide

so limiting agent is p-nitroacetanilide

B)percent yeild

they isolated 100 grams of p-nitroaniline

no of moles of p-nitroaniline = weight of p-nitroaniline / molar mass of p-nitroaniline = 100 g / 138.13 g /mol =

=0.724 mol

formula for the percentage of yield is

% of yield = (moles of products / moles of reactants ) x 100

= (0.724 / 1) x 100

= 72.4 %