Answer:
[tex]\large \boxed{\text{0.012 mol}}[/tex]
Explanation:
We will need a balanced equation with moles, so let's gather all the information in one place.
CH₃C₆H₄NH₂·HCl + (CH₃CO)₂O ⟶ CH₃C₆H₄NHCOCH₃ + junk
V/mL: 70.
c/mol·L⁻¹: 0.167
For simplicity in writing , let's call p-toluidine hydrochloride A and N-acetyl-p-toluidine B.
The equation is then
A + Ac₂O ⟶ B + junk
1. Moles of A
[tex]\text{Moles of A} = \text{70. mL A}\times \dfrac{\text{0.167 mmol A}}{\text{1 mL A}}= \text{12 mmol A}[/tex]
2. Moles of B
The molar ratio is 1 mol B:1 mol A
Moles of B = moles of A = 12 mmol = 0.012 mol
[tex]\text{You can prepare $\large \boxed{\textbf{0.012 mol}}$ of N-acetyl-p-toluidine. }[/tex]
