aspirin is synthesized through the following reaction
C7H603+C4H6O3=C9H8O4=HC2H3O2
a)Assuming the maximum yield that could be obtained is 76.8% what mass of Aspirin C9H8O4 could be obtained from this reaction when 2.00 * 10^2 grams of C7H6O3 reacts with 145.30 grams of C4H6O3.
b)calculate the mass of excess reagent that is left over in this reaction?
Let's call these r1 and r2 for reagent 1 and reagent 2. So r1 is C7H6O3 and r2 is C4H6O3. You made a typo in the equation. You typed an = sign for a + sign.
C7H603+C4H6O3=C9H8O4+HC2H3O2
a. Which reactant is limiting?
mols r1 = grams/molar mass = estimated 1.45
mols r2 = g/molar mass = approx 1.42
This is a 1:1 reaction ratio; therefore, the smaller number is the limiting reagent and the other is the excess reagent. Limiting reagent is r2.
Therefore, 1.42 mols product will be formed. Convert to grams. g = mols x molar mass product. Multiply by percent yield (0.768) to obtain the actual yield.
b. r2 is the limiting reagent. r1 is the excess. How much of the excess reagent is used. That is mols r2 x (1 mol r1/1 mol r2) = 1.42 x (1/1) = 1.42 used. You had 1.45 initially. The difference is what was not reacted. That x molar mass = grams.
To answer these questions, we need to follow a few steps:
Step 1: Determine the limiting reagent.
Step 2: Calculate the theoretical yield.
Step 3: Calculate the mass of excess reagent.
a) Let's start with determining the limiting reagent.
1 mole of C7H6O3 reacts with 1 mole of C4H6O3 to produce 1 mole of C9H8O4.
Given:
Mass of C7H6O3 = 2.00 * 10^2 grams
Mass of C4H6O3 = 145.30 grams
To determine the limiting reagent, we need to calculate the moles of each reactant.
Molar mass of C7H6O3 = 138.12 g/mol
Molar mass of C4H6O3 = 102.09 g/mol
Moles of C7H6O3 = mass / molar mass
= 200 / 138.12
= 1.447 mol
Moles of C4H6O3 = mass / molar mass
= 145.30 / 102.09
= 1.424 mol
Since both reactants have a 1:1 molar ratio with the product, the limiting reagent is the one that produces the fewer moles. In this case, C4H6O3 is the limiting reagent.
b) Now, let's calculate the theoretical yield of Aspirin (C9H8O4) using the limiting reagent.
Moles of C9H8O4 = moles of C4H6O3
= 1.424 mol
Molar mass of C9H8O4 = 180.16 g/mol
Mass of Aspirin (C9H8O4) = moles of C9H8O4 * molar mass of C9H8O4
= 1.424 * 180.16
= 257.24 grams
However, the maximum yield that could be obtained is given as 76.8%. Therefore, we need to adjust the calculated mass accordingly.
Adjusted mass of Aspirin (C9H8O4) = theoretical yield * maximum yield percentage
= 257.24 * (76.8/100)
= 197.66 grams
Therefore, the maximum mass of Aspirin (C9H8O4) that could be obtained from this reaction is 197.66 grams.
Now, let's move on to calculating the mass of excess reagent.
Mass of excess reagent = initial mass - mass reacted
The initial mass of C7H6O3 = 200 grams
Mass reacted = moles of limiting reagent * molar mass of C7H6O3
Moles of limiting reagent = moles of C4H6O3
= 1.424 mol
Mass reacted = moles of limiting reagent * molar mass of C7H6O3
= 1.424 * 138.12
= 196.13 grams
Mass of excess reagent = initial mass - mass reacted
= 200 - 196.13
= 3.87 grams
Therefore, the mass of excess reagent remaining in this reaction would be 3.87 grams.