C2H5OH(l)+CH3COOH(l) → CH3COOC2H5(l)+H2O(l) If the yield of ethyl ethanoate obtained when 20.00 g of ethanol is reacted with excess ethanoic acid is 30.27 g, calculate the percentage yield

The percentage yield is 79.12%

Explanation:

The first thing we need to calculate here is the theoretical yield of ethyl ethanoate that can be obtained from 20g of ethanol.

Since the mole ratio is 1:1, then this is quite straightforward.

What we need to obtain at first here is the number of theoretical moles of ethanol reacting.

That would be mass of ethanol/molar mass of ethanol

Molar mass of ethanol is 46g/mol

Thus the number of moles is 20/46 = 0.4348 mole

Like it is indicated earlier, since the number of moles are equal from the balanced equation, it also means that 0.4348 mole of ethylethanoate is produced

Now, we need to know the mass of ethyl ethanoate produced

The mass can be calculated by multiplying the number of moles by the molar mass

The molar mass of ethyl ethanoate is 88g/mol

So the mass of it produced = 0.4348 * 88 = 38.2624 or let’s just say 38.26

Thus, the percentage yield will be;

Actual yield/Theoretical yield * 100%

From the question, our actual yield is 30.27g while our calculated theoretical yield is 38.26g

= 30.27/38.26 * 100% = 79.12%

h let me think