If 11.53g hydrogen sulfide reacts with 3.97g oxygen gas in the following reaction:

2H2S + O2 + H2O

which is the maximum mass of sulfur that can be produced?

which is the limiting reactant ?

which reactant is in excess ? by how many grams?

To find the maximum mass of sulfur that can be produced, we first need to determine the limiting reactant.

1) Calculate the number of moles for each reactant:
- Hydrogen sulfide (H2S): 11.53g / molar mass of H2S = x moles
- Oxygen gas (O2): 3.97g / molar mass of O2 = y moles

Note: The molar mass of H2S is approximately 34.08g/mol, and the molar mass of O2 is approximately 32.00g/mol.

2) Compare the mole ratios of the reactants to the balanced equation. From the balanced equation, we can see that the ratio of H2S to O2 is 2:1. We need to determine which reactant has the lower mole ratio.

- Calculate the mole ratio of H2S to O2: x moles H2S / y moles O2
- Calculate the mole ratio of O2 to H2S: y moles O2 / x moles H2S

3) Compare the mole ratios obtained in step 2 to determine the limiting reactant. The limiting reactant is the one with the smaller mole ratio. This indicates which reactant is fully consumed and determines the maximum yield of the product.

4) Once the limiting reactant is determined, we use the stoichiometry given in the balanced equation to calculate the mass of sulfur produced. From the balanced equation, we can see that the mole ratio of H2S to sulfur is 2:1.

- Calculate the moles of sulfur produced: x moles H2S x (1 mole sulfur / 2 moles H2S)
- Calculate the mass of sulfur produced: moles of sulfur produced x molar mass of sulfur

5) To find out which reactant is in excess, we subtract the moles or grams of the limiting reactant from the original amount of the reactant used. The excess amount indicates how much of the reactant is left over after the reaction is complete.

By following these steps, we can determine the maximum mass of sulfur, the limiting reactant, and the excess reactant.