Respuesta :
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:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Molarity of p-toluidine hydrochloride solution = 0.167 M
Volume of solution = 70. mL
Putting values in above equation, we get:
[tex]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[/tex]
The chemical equation for the reaction of p-toluidine hydrochloride and acetic anhydride follows:
[tex]\text{p-toluidine hydrochloride}+\text{Acetic anhydride}\rightarrow \text{N-acetyl-p-toluidine}[/tex]
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 = [tex]\frac{1}{1}\times 0.0117=0.0117mol[/tex] of N-acetyl-p-toluidine
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
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:
[tex]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[/tex]
Hence, the maximum amount of N-acetyl-p-toluidine that can be prepared is 1.7 grams.
