Question

use the numbers from image 2 to help me answer all parts from image 1. I am confused how to solve. will rate well

1. The molecular weight of vanillin (a solid) is 152.15. The molecular weight of acetic anhydride (a liquid) is 102.04 and its density is 1.082 g/mL. How many mmoles of vanillin and how many mmoles of acetic anhydride are being reacted in the BASE CATALYZED reaction as described in the textbook? a. 1.97 mmol of vanillin and 8.45 mmol acetic anhydride b. 45.6 mmol of vanillin and 82 mmol acetic anhydride c. 8.45 mmol of vanillin and 2 mmol acetic anhydride d. 2 mmol of vanillin and 2 mmol acetic anhydride Answer: 2. Limiting reagent (Based on your answer to the previous question): Answer: 3. Theoretical yield in moles and grams (or mg) based on your answer to the previous question: Answer:

Answer 1

$\begin{aligned}&C_8H_8O_3\;\;\;\;\;+&C_4H_6O_3\;\;\rightleftharpoons \;\;&C_{10}H_{10}O_4\\&1.97\;mmol &8.45\;mmol\;\;\;\;\;&1.97\;mmol\end{aligned}$

1. No. of moles of vaniline  

          

0.3 g 152.15 g/mol = 1.97 mmol

   No. of moles of acetic anhydride

  $\begin{aligned} &= \mathrm{\frac{density\times volume}{M_{wt}}}\\&= \frac{1.082\;g/mL\times0.8\;mL}{102.04}\\&=8.45\;mmol\end{aligned}$

answer : option (a)

2. Limiting reagent = moles of vaniline

= 1.97 mmol

3. Theoretical yield = 1.97 mmol

$\begin{aligned}&= \mathrm{\frac{1.97\;mmol}{194.15\;g/mol\;(M_{wt}=C_{10}H_{10}O_4)}}\\&=0.38\;g\end{aligned}$