What is the theoretical yield of H2S

Best Answer: Whenever you do limiting reagent questions, or stoichiometry in general, everything must be converted to moles in order to compare them. So if you were given masses, you'd need to convert them to moles anyway. The fact that you are given moles instead of mass actually cuts one step out of your calculations.

Start by picking one of the reagents,

Work out how much of the other reagent is needed to react all of the reagent you picked.

I am going to pick the 4 moles of Al2S3

Let's look at the balanced equation, it tells you that

1 mole of Al2S3 needs 6 moles of H2O to fully react

So, to fully react 4 moles of Al2S3 you will need 6 x as much H2O

moles H2O needed = 4 mol Al2S3 x (6 mol H2O / 1 mol Al2S3)

= 24 moles H2O

BUT, you are only given 4 moles of H2O. So you have no where near enough H2O to react all your 4 moles of Al2S3. Thus H2O is the limiting reagent.

If you had picked the 4 moles of H2O you would work out how many moles of Al2S3 are needed to fully react all the 4 moles of H2O

From the balanced equation ...

6 moles H2O reacts with 1 mole Al2S3

Thus moles Al2S3 needed = 4 mol H2O x (1 mol Al2S3 / 6 mol H2O)

= 0.67 moles Al2S3 needed

Since you are given 4 moles of Al2S3, and only need 0.67 moles the Al2S3 is in excess, and so H2O is limiting.

The theoretical yield is the maximum amount of product possible given the amounts of each reagent provided. It occurs if all the limiting reagent fully reacts.

So now work out how many moles of H2S you'd get from 4 moles of H2O

6 moles H2O react to produce 3 moles H2S

moles H2S possible = 4.0 mole H2O x (3 mol H2S / 6 mol H2O)

= 2 mol H2S

the theoretical yield of H2S = 2.0 mol