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Zinaida [17]
3 years ago
6

Aspirin sun thesis Green Chemistry and Assime the aspirin is prepared by the following reaction and that 10.09. of salicylic aci

d and an excess of acetic anhydride are used. If you get 5.og of aspirin: Cy HCO3 +Cu the Oz - Cats O4 + C a Hu Oz what is the percent yield of aspirin? show calculations Formula : Experimental performance Theoretical performance X 100
Chemistry
1 answer:
klemol [59]3 years ago
6 0

<u>Answer:</u> The percentage yield of aspirin is 38.02 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For salicylic acid:</u>

Given mass of salicylic acid (C_7H_6O_3) = 10.09 g

Molar mass of salicylic acid (C_7H_6O_3) = 138.12 g/mol

Putting values in equation 1, we get:

\text{Moles of salicylic acid}=\frac{10.09g}{138.12g/mol}=0.0730mol

The chemical equation for the formation of aspirin follows:

C_7H_6O_3+C_4H_6O_3\rightarrow C_9H_8O_4+CH_3COOH

As, acetic anhydride is present in excess. So, it is considered as an excess reagent.

Thus, salicylic acid is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of salicylic acid produces 1 mole of aspirin.

So, 0.0730 moles of salicylic acid will produce = \frac{1}{1}\times 0.0730=0.0730mol of aspirin

Now, calculating the mass of aspirin from equation 1, we get:

Molar mass of aspirin = 180.16 g/mol

Moles of aspirin = 0.073 moles

Putting values in equation 1, we get:

0.073mol=\frac{\text{Mass of aspirin}}{180.16g/mol}\\\\\text{Mass of aspirin}=13.15g

To calculate the percentage yield of aspirin, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of aspirin = 5.0 g

Theoretical yield of aspirin = 13.15 g

Putting values in above equation, we get:

\%\text{ yield of aspirin}=\frac{5.0g}{13.15g}\times 100\\\\\% \text{yield of aspirin}=38.02\%

Hence, the percent yield of aspirin is 38.01 %.

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Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.
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Explanation:

<u></u>

<u>1. Value of [Ag⁺]  in a saturated solution of AgCl in distilled water.</u>

The value of [Ag⁺]  in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

The ICE (initial, change, equilibrium) table is:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                    0

C          -s                      +s                  +s

E         X - s                   s                     s

Since s is very small, X - s is practically equal to X and is a constant, due to which the concentration of the solids do not appear in the Ksp equation.

Thus, the Ksp equation is:

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  • Ksp = s × s
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By substitution:

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Rounding to two significant figures:

  • [Ag⁺] = 1.3 × 10⁻⁵M ← answer

<u></u>

<u>2. Molar solubility of AgCl(s) in seawater</u>

Since, the conentration of Cl⁻ in seawater is 0.54 M you must introduce this as the initial concentration in the ECE table.

The new ICE table will be:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                  0.54

C          -s                      +s                  +s

E         X - s                   s                     s + 0.54

The new equation for the Ksp equation will be:

  • Ksp = [Ag⁺] [Cl⁻]
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  • Ksp = s² + 0.54s

By substitution:

  • 1.8 × 10⁻¹⁰ = s² + 0.54s
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Now you must solve a quadratic equation.

Use the quadratic formula:

     

     s=\dfrac{-0.54\pm\sqrt{0.54^2-4(1)(-1.8\times 10^{-10})}}{2(1)}

The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer

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In the case of solubility reactions, this is known as the common ion effect: when the solution contains one of the ions that is formed by the solid reactant, the reaction will proceed in less proportion, i.e. less reactant can be dissolved.

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