Question

Aspirin (C9H8O4, molecular mass 180.2 g/mol) is synthesized by the reaction of salicylic acid (C7H6O3, molecular mass 138.1 g/mol) with acetic anhydride (C4H6O3, molecular mass 102.1 g/mol) according to the reaction 2 C7H6O3 (s) + C4H6O3(l) → 2C9H8O4(s)+ H2O(l) 2.0g of salicylic acid and 5.4 g of acetic anhydride are mixed and the reaction is allowed to go to completion. Do each of the following: A. Identify the limiting reagent. B. State how many grams of the excess reagent remain. C. State how many grams of product are produced.

Answer 1

Answer:

A. The Salicylic acid $C_{7}H_{6}O_{3}$ is the limiting reagent.

B. 4.66 g of the acetic anhydride $C_{4}H_{6}O_{3}$ remain.

C. 2.6 g of $C_{9}H_{8}O_{4}$ are produced.

Explanation:

A. First write the balanced chemical equation fior the asprin synthesis:

$_{2}C_{7}H_{6}O_{3}_{(s)}+C_{4}H_{6}O_{3}_{(l)}=_{2}_{(s)}+H_{2}O_{(l)}$

Then determine the limiting reagent.

To determine the limiting reagent first divide the mass of each compound between its molar mass, then divide this quantity by the number of moles given by the reaction and the smallest number will be the limiting reagent.

- For the Salicylic acid $C_{7}H_{6}O_{3}$:

$2.0g*\frac{1mol}{138.1g}=0.014$

$\frac{0.014}{2}=0.007$

- For the Acetic anhydride $C_{4}H_{6}O_{3}$:

$5.4g*\frac{1mol}{102.1g}=0.053$

$\frac{0.052}{1}=0.053$

The smallest number is for the salicylic acid therefore it is the limiting reagent.

B. Calculate how many grams of the excess reagent remain.

- First calculate how many grams of acetic anhydride reacts:

$2.0gC_{7}H_{6}O_{3}*\frac{1molC_{7}H_{6}O_{3}}{138.1gC_{7}H_{6}O_{3}}*\frac{1molC_{7}H_{6}O_{3}}{2molesC_{7}H_{6}O_{3}}*\frac{102.1gC_{4}H_{6}O_{3}}{1molC_{4}H_{6}O_{3}}=0.74gC_{4}H_{6}O_{3}$ reacts

- Then subtract the quantity of acetic anhydride that reacts from the quantity that are mixed:

$5.4gC_{4}H_{6}O_{3}-0.74gC_{4}H_{6}O_{3}=4.66gC_{4}H_{6}O_{3}$ remain

C. Calculate how many grams of product are produced:

$2.0gC_{7}H_{6}O_{3}*\frac{1molC_{7}H_{6}O_{3}}{138.1gC_{7}H_{6}O_{3}}*\frac{2molesC_{9}H_{8}O_{4}}{2molesC_{7}H_{6}O_{3}}*\frac{180.2gC_{9}H_{8}O_{4}}{1molC_{9}H_{8}O_{4}}=2.6gC_{9}H_{8}O_{4}$ are produced.