# Question #6e6f8

Dec 5, 2016

The minimum volume of acetic anhydride is 1.49 mL.

#### Explanation:

I'll write the molar masses above the formulas for easy access.

${M}_{r} : \textcolor{w h i t e}{l l} 138.12 \textcolor{w h i t e}{m m m l} 102.09$
$\textcolor{w h i t e}{m m l} \underbrace{\text{C"_7"H"_6"O"_3)_color(red)("salicylic acid") + underbrace("C"_4"H"_6"O"_3)_color(red)("acetic anhydride") → underbrace("C"_9"H"_8"O"_4)_color(red)("aspirin") + underbrace("C"_2"H"_4"O"_2)_color(red)("acetic acid}}$

Now let's rewrite the equation for easier typing.

$\text{Sal" + "Ac"_2"O" → "Asp" + "HOAc}$

Step 2. Calculate the moles of $\text{Sal}$.

$\text{Moles of Sal" = 2.182 color(red)(cancel(color(black)("g Sal"))) × "1 mol Sal"/(138.12 color(red)(cancel(color(black)("g Sal")))) = "0.015 798 mol Sal}$

Step 3. Calculate the moles of $\text{Ac"_2"O}$.

$\text{Moles of Ac"_2"O" = "0.015 798" color(red)(cancel(color(black)("mol Sal"))) × ("1 mol Ac"_2"O")/(1 color(red)(cancel(color(black)("mol Sal")))) = "0.015 798 mol Ac"_2"O}$

Step 4. Calculate the mass of $\text{Ac"_2"O}$.

$\text{Mass of Ac"_2"O" = "0.015 798" color(red)(cancel(color(black)("mol Ac"_2"O"))) × ("102.09 g Ac"_2"O")/(1 color(red)(cancel(color(black)("mol Ac"_2"O")))) = "1.6128 g Ac"_2"O}$

Step 5. Calculate the volume of $\text{Ac"_2"O}$.

$\text{Volume of Ac"_2"O" = 1.6128 color(red)(cancel(color(black)("g Ac"_2"O"))) × ("1 mL Ac"_2"O")/(1.08 color(red)(cancel(color(black)("ml Ac"_2"O")))) = "1.49 mL Ac"_2"O}$