We read your photo as:

A) Which is the limiting reactant? Given the density of acetic anhydride is 1.030 g/ml? b) What is the percent yield of this reaction?

Chemistry

ScanSolve AI Answer

Answers

2 parts
a

Salicylic acid\text{Salicylic acid}

b

81.72%81.72\%

Step-by-step solution

6 steps

To solve this problem, we first need to determine the molar masses of the reactants and products, then calculate the moles of each reactant to find the limiting reactant, and finally use the limiting reactant to calculate the theoretical yield and percent yield.

The balanced chemical equation is:

C7H6O3+(CH3CO)2O→C9H8O4+CH3COOH\text{C}_7\text{H}_6\text{O}_3 + \text{(CH}_3\text{CO)}_2\text{O} \rightarrow \text{C}_9\text{H}_8\text{O}_4 + \text{CH}_3\text{COOH}

(Salicylic acid) + (Acetic anhydride) →\rightarrow (Aspirin) + (Acetic acid)

Molar masses:

  • Salicylic acid (C7H6O3\text{C}_7\text{H}_6\text{O}_3):
(7×12.01)+(6×1.008)+(3×16.00)=84.07+6.048+48.00=138.118 g/mol(7 \times 12.01) + (6 \times 1.008) + (3 \times 16.00) = 84.07 + 6.048 + 48.00 = 138.118 \text{ g/mol}
  • Acetic anhydride ((CH3CO)2O\text{(CH}_3\text{CO)}_2\text{O} or C4H6O3\text{C}_4\text{H}_6\text{O}_3):
(4×12.01)+(6×1.008)+(3×16.00)=48.04+6.048+48.00=102.088 g/mol(4 \times 12.01) + (6 \times 1.008) + (3 \times 16.00) = 48.04 + 6.048 + 48.00 = 102.088 \text{ g/mol}
  • Aspirin (C9H8O4\text{C}_9\text{H}_8\text{O}_4):
(9×12.01)+(8×1.008)+(4×16.00)=108.09+8.064+64.00=180.154 g/mol(9 \times 12.01) + (8 \times 1.008) + (4 \times 16.00) = 108.09 + 8.064 + 64.00 = 180.154 \text{ g/mol}

Part a· 4 steps

Which is the limiting reactant?

  1. Calculate the moles of salicylic acid.

    Moles of salicylic acid=massmolar mass=3.0779 g138.118 g/mol=0.022284 mol\text{Moles of salicylic acid} = \frac{\text{mass}}{\text{molar mass}} = \frac{3.0779 \text{ g}}{138.118 \text{ g/mol}} = 0.022284 \text{ mol}
  2. Calculate the mass of acetic anhydride using its density and volume.

    Mass of acetic anhydride=density×volume=1.080 g/mL×5.050 mL=5.454 g\text{Mass of acetic anhydride} = \text{density} \times \text{volume} = 1.080 \text{ g/mL} \times 5.050 \text{ mL} = 5.454 \text{ g}
  3. Calculate the moles of acetic anhydride.

    Moles of acetic anhydride=massmolar mass=5.454 g102.088 g/mol=0.053426 mol\text{Moles of acetic anhydride} = \frac{\text{mass}}{\text{molar mass}} = \frac{5.454 \text{ g}}{102.088 \text{ g/mol}} = 0.053426 \text{ mol}
  4. Determine the limiting reactant.

    From the balanced equation, the stoichiometric ratio between salicylic acid and acetic anhydride is 1:1.
    Since 0.022284 mol0.022284 \text{ mol} (salicylic acid) is less than 0.053426 mol0.053426 \text{ mol} (acetic anhydride), salicylic acid is the limiting reactant.

    The limiting reactant is salicylic acid.

    Salicylic acid\boxed{\text{Salicylic acid}}

Part b

81.72%81.72\%

What is the percent yield of this reaction?

  1. Calculate the theoretical yield of aspirin.

    Based on the balanced equation, 1 mole of salicylic acid produces 1 mole of aspirin.

    Moles of aspirin (theoretical)=moles of limiting reactant=0.022284 mol\text{Moles of aspirin (theoretical)} = \text{moles of limiting reactant} = 0.022284 \text{ mol} Theoretical yield of aspirin=moles×molar mass=0.022284 mol×180.154 g/mol=4.0148 g\text{Theoretical yield of aspirin} = \text{moles} \times \text{molar mass} = 0.022284 \text{ mol} \times 180.154 \text{ g/mol} = 4.0148 \text{ g}
  2. Calculate the percent yield.

    Percent yield=actual yieldtheoretical yield×100%\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% Percent yield=3.281 g4.0148 g×100%=81.72%\text{Percent yield} = \frac{3.281 \text{ g}}{4.0148 \text{ g}} \times 100\% = 81.72\%

    The percent yield of this reaction is 81.72%.

    81.72%\boxed{81.72\%}
Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp