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lozanna [386]
3 years ago
13

What is the maximum number of moles of N-acetyl-p-toluidine can be prepared from 70. milliliters of 0.167 M p-toluidine hydrochl

oride and an excess of acetic anhydride in an acetate buffer? Enter only the number with two significant figures.
Chemistry
1 answer:
Kitty [74]3 years ago
3 0

Answer:

\large \boxed{\text{0.012 mol}}  

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-<em>p</em>-toluidine B.

The equation is then

A + Ac₂O ⟶ B + junk

1. Moles of A

\text{Moles of A} = \text{70. mL A}\times \dfrac{\text{0.167 mmol A}}{\text{1 mL A}}= \text{12 mmol A}

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

\text{You can prepare $\large \boxed{\textbf{0.012 mol}}$ of N-acetyl-p-toluidine. }

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approximately 15.1 grams.

Explanation:

The key to chemistry is to change everything to moles. Then when you have the answer in moles change the answer back to grams, liters, or whatever you want.

change 25 grams of potassium chlorate to moles.

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