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1) Aspirin (C 9 H 8 O 4 ) is synthesized by mixing salicylic acid (C 7 H 6 O 3 )

ID: 792290 • Letter: 1

Question

1) Aspirin (C9H8O4) is synthesized by mixing salicylic acid (C7H6O3) with excess acetic anhydride (C4H6O3). The balanced equation is:

C7H6O3 + C4H6O3 ------------> C9H8O4 + C2H4O2

Molar mass/ g.mol^-1

C2H4O2            60

C4H6O3            102

C7H6O3            138

C9H8O4             180

When 2.0g of salicylic acid is mixed with excess acetic anhydrude, 1.7g of aspirin is collected . What is the percent yield for this reaction?

2) Starting with 4.3 L if 0.58 M Na2CO3(aq) what volume of this solution is needed to prepare 250 mL of 0.20 M Na2CO3(aq) solution?

3) If 26.32 mL of .100 M H2SO4 is exactly neutralized by 34.56 of NaOH solution, what is the molar concentration?

4) A solution is prepared by dissolving 3.422g of Al2(SO4)3, molar mass 342.2 g.mol^-1, in water to obtain 0.100 L of solution. What is the molar concentration of sulfate ion in this solution?

Please explain! Step by step if possible! Thank you!

Explanation / Answer

1) by the given reaction


C7H6O3 + C4H6O3 ------------> C9H8O4 + C2H4O2


1 mole of salicylic acid should give 1 mole of aspirin


that is 138 g of salicylic acid should give 180 g of aspirin


let 2 g of salicylic acid give x g of aspirin

then x= 2 *180/138 = 2.61 g


so 2 g of salicylic acid should give 2.61 go aspirin


but give only 1.7 g aspirin is obtained


so percentage yield= 1.7/ 2.61 x100

percentage yield is 63.13 %


2) Final requirement is 250 mL of 0.20 M of Na2CO3 solution


moles = molarity x volume /1000

= 0.20 x 250/1000

moles=0.05



these moles should come from 0.58 M Na2C03


so conc=moles x 1000/volume

0.58= 0.05x1000/volume

volume= 86.02 ml



3)

2NaOH + H2SO4 -> Na2SO4 + 2H20

for H2SO4 Normality = molarity x 2

consider N1V1=N2V2

that is 0.2 x 26.32 = 34.56 x N2

N2= 0.15 N

for NaOH molarity = normality

Molarity = O.15 M


4) Given mass=3.422 g and M.w = 342.2 g and volume=0.1 L


conc= mass /M.W x Volume(L)

= 3.422 /342.2 x0.1

conc= 0.1 M