Question

What is the maximum number of grams of N-acetyl-p-toluidine can be prepared from 70. milliliters of 0.167 M p-toluidine hydrochloride and an excess of acetic anhydride in an acetate buffer

Answer 1

Answer: The maximum amount of N-acetyl-p-toluidine that can be prepared is 1.7 grams.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of p-toluidine hydrochloride solution = 0.167 M

Volume of solution = 70. mL

Putting values in above equation, we get:

$0.167M=\frac{\text{Moles of p-toluidine hydrochloride}\times 1000}{70}\\\\\text{Moles of p-toluidine hydrochloride}=\frac{0.167\times 70}{1000}=0.0117mol$

The chemical equation for the reaction of p-toluidine hydrochloride and acetic anhydride follows:

$\text{p-toluidine hydrochloride}+\text{Acetic anhydride}\rightarrow \text{N-acetyl-p-toluidine}$

By Stoichiometry of the reaction:

1 mole of p-toluidine hydrochloride produces 1 mole of N-acetyl-p-toluidine

So, 0.0117 moles of p-toluidine hydrochloride will produce = $\frac{1}{1}\times 0.0117=0.0117mol$ of N-acetyl-p-toluidine

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of N-acetyl-p-toluidine = 149.2 g/mol

Moles of N-acetyl-p-toluidine = 0.0117 moles

Putting values in equation 1, we get:

$0.0117mol=\frac{\text{Mass of N-acetyl-p-toluidine}}{149.2g/mol}\\\\\text{Mass of N-acetyl-p-toluidine}=(0.0117g/mol\times 149.2)=1.7g$

Hence, the maximum amount of N-acetyl-p-toluidine that can be prepared is 1.7 grams.