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.

Step-by-step explanation:

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

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 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}* 1000}{70}\\\\\text{Moles of p-toluidine hydrochloride}=(0.167* 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 =
`(1)/(1)* 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* 149.2)=1.7g`

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