Chemistry, 31.01.2020 00:51 Aylin042277

What is the theoretical yield of h2s

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Best 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

Answer from: Quest

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

Answer from: Quest

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