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Aspirin can be made in the laboratory by reacting acetic anhydride (C4HgO3) with
Chapter 8, Problem 91P(choose chapter or problem)
Aspirin can be made in the laboratory by reacting acetic anhydride (\(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}\)) with salicylic acid (\(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\)) to form aspirin (\(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\)) and acetic acid (\(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\)). The balanced equation is:
\(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}+\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3} \rightarrow \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}+\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\)
In a laboratory synthesis, a student begins with 5.00 mL of acetic anhydride (density = 1.08 g/mL) and 2.08 g of salicylic acid. Once the reaction is complete, the student collects 2.01 g of aspirin. Determine the limiting reactant, theoretical yield of aspirin, and percent yield for the reaction.
Equation Transcription:
Text Transcription:
C_4 H_6 O_3
C_7 H_6 O_3
C_9 H_8 O_4
C_2 H_4 O_2
C_4 H_6 O_3 + C_7 H_6 O_3 right arrow C_9 H_8 O_4 + C_2 H_4 O_2
Questions & Answers
QUESTION:
Aspirin can be made in the laboratory by reacting acetic anhydride (\(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}\)) with salicylic acid (\(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\)) to form aspirin (\(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\)) and acetic acid (\(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\)). The balanced equation is:
\(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}+\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3} \rightarrow \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}+\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\)
In a laboratory synthesis, a student begins with 5.00 mL of acetic anhydride (density = 1.08 g/mL) and 2.08 g of salicylic acid. Once the reaction is complete, the student collects 2.01 g of aspirin. Determine the limiting reactant, theoretical yield of aspirin, and percent yield for the reaction.
Equation Transcription:
Text Transcription:
C_4 H_6 O_3
C_7 H_6 O_3
C_9 H_8 O_4
C_2 H_4 O_2
C_4 H_6 O_3 + C_7 H_6 O_3 right arrow C_9 H_8 O_4 + C_2 H_4 O_2
ANSWER:
Step 1 of 5
Given that initial volume of acetic anhydride is 5.00 mL and density is 1.08 g/mL
Mass = density volume = 5.00 mL 1.08 g/mL=5.4 g
Number of moles of = Given mass/Molecular mass
Molecular mass of = 102g/mol
Thus number of moles of = = 0.052 moles