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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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What is the molarity of solution obtained when 5.71 g of sodium carbonate-10-water is dissolved in water and made up to 250.0 cm
disa [49]

We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.

The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O)  is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³

The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.

Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.

5.71 grams of Na₂CO₃.10 H₂O is equal to \frac{5.71}{286}= 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.

Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is \frac{0.0199 X 1000}{250} = 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³


3 0
3 years ago
Calculate how many grams of the first reactant are necessary to completely react with 17.3 g of the second reactant. the reactio
soldier1979 [14.2K]

Taking into account the reaction stoichiometry, 16.611 grams of Na₂CO₃ are necessary to completely react with 17.3 g of CuCl₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Na₂CO₃ + CuCl₂  → CuCO₃ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Na₂CO₃: 1 mole
  • CuCl₂: 1 mole
  • CuCO₃: 1 mole
  • NaCl: 2 moles

The molar mass of the compounds is:

  • Na₂CO₃: 129 g/mole
  • CuCl₂: 134.45 g/mole
  • CuCO₃: 123.55 g/mole
  • NaCl: 58.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Na₂CO₃: 1 mole ×129 g/mole= 129 grams
  • CuCl₂: 1 mole ×134.45 g/mole= 134.45 grams
  • CuCO₃: 1 mole ×123.55 g/mole= 123.55 grams
  • NaCl: 2 mole ×58.45 g/mole=116.9 grams

<h3>Mass of CuCl₂ required</h3>

The following rule of three can be applied: If by reaction stoichiometry 134.35 grams of CuCl₂ react with 129 grams of Na₂CO₃, 17.3 grams of CuCl₂ react with how much mass of Na₂CO₃?

mass of Na₂CO₃= (17.3 grams of CuCl₂× 129 grams of Na₂CO₃)÷ 134.35 grams of CuCl₂

<u><em>mass of Na₂CO₃= 16.611 grams</em></u>

Finally, 16.611 grams of Na₂CO₃ is required.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

6 0
10 months ago
PLEASE HELP
DENIUS [597]

I believe that it is B

3 0
2 years ago
How much heat is needed to change the temperature of 5g of water from 20 oC to 37 oC?
Dafna1 [17]

Answer:

The answer to your question is Q = 355.64 J

Explanation:

Data

Heat = Q = ?

Temperature 1 = T1 = 20°C

Temperature 2 = T2 = 37°C

mass = m = 5 g

Specific heat = Cp = 4.184 J/g°C

Formula

Q = mCp(T2 - T1)

-Substitution

Q = (5)(4.184)(37 - 20)

-Simplification

Q = (5)(4.184)(17)

-Result

Q = 355.64 J

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2 years ago
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A _____ is the smallest particle of a substance that has all the properties of the substance. neutron cell atom molecule
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