Determine the # of moles of Al2S3 that can be prepared by the reaction of 0.400 moles of Aluminum with 0.700 moles of Sulfur, and Also find the # of moles of reactant in excess

See the response to your question before this. I solve these limiting reagent problems (you know it is a limiting reagent problem when BOTH reactants are given) by solving TWO separate simple stoichiometry problems as in the previous example. You will obtain two answers, of course, but the correct answer in limiting reagent problems is ALWAYS the smaller one. The amount of reagent in excess is worked as a simple stoichiometry problem, too. Post your work if you get stuck.

To determine the number of moles of Al2S3 that can be prepared, we first need to identify the limiting reagent. The limiting reagent is the reactant that is completely consumed in a chemical reaction and determines the amount of product formed.

To find the limiting reagent, we compare the stoichiometric ratio between Aluminum (Al) and Al2S3, and between Sulfur (S) and Al2S3. The balanced equation for the reaction is:

2 Al + 3 S → Al2S3

The stoichiometric ratio tells us that 2 moles of Aluminum react with 3 moles of Sulfur to produce 1 mole of Al2S3.

Given:
- Moles of Aluminum (Al) = 0.400 moles
- Moles of Sulfur (S) = 0.700 moles

To determine the limiting reagent, we need to calculate the number of moles of Al2S3 that can be formed from each reactant. We will use the stoichiometric ratios mentioned earlier.

1) Calculate the number of moles of Al2S3 that can be formed from Aluminum (Al):
Moles of Al2S3 = (Moles of Al/2) = 0.400 moles/2 = 0.200 moles

2) Calculate the number of moles of Al2S3 that can be formed from Sulfur (S):
Moles of Al2S3 = (Moles of S/3) = 0.700 moles/3 = 0.233 moles (approximately)

We see that we can only form 0.200 moles of Al2S3 from 0.400 moles of Aluminum, while we can form approximately 0.233 moles of Al2S3 from 0.700 moles of Sulfur.

Since we can form a lesser amount of Al2S3 from Aluminum, Aluminum is the limiting reagent. Therefore, the number of moles of Al2S3 that can be prepared is 0.200 moles.

To find the number of moles of the reactant in excess, we subtract the number of moles of the limiting reagent from the respective reactant.

Moles of reactant in excess = Moles of reactant - Moles used in the reaction

For Sulfur (S):
Moles of reactant in excess = (Moles of S - Moles used in the reaction)
= (0.700 moles - 0.233 moles)
= 0.467 moles (approximately)

Therefore, the number of moles of Sulfur in excess is approximately 0.467 moles.