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What is the maximum number of moles 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? Enter only the number with two significant figures.

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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⁻Âč:             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]

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