Chemistry homework

Question

According to the following reaction, how many grams of water will be formed upon the complete reaction of 28.0 grams of oxygen gas with excess hydrogen sulfide?

hydrogen sulfide (g) + oxygen (g) water (l) + sulfur dioxide (g)

Answer 1

From your previous question, and my previous answer to your previous question, we have the following:

Let's turn the word equation into a chemical equation and balance it.

hydrogen sulfide (g) = H₂S(g)

oxygen (g) = O₂(g)

water (l) = H₂O(l)

sulfur dioxide = SO₂(g)

2H₂S(g) + 3O₂(g) ==> 2H₂O(l) + 2SO₂(g) ... balanced equation

From the balanced equation we see TWO moles of H₂O are formed from THREE moles of O₂ when H₂S is present in excess. Using this mole ratio and the fact that we have 28.0 g of O₂, we proceed as follows:

28.0 g O₂ x 1 mol O₂ / 32 g O₂ = 0.875 mols O₂ present

0.875 mols O₂ x 2 mols H₂O / 3 mols O₂ = 0.583 mols H₂O formed

0.583 mols H₂O x 18 g H₂O / mol H₂O = 10.5 g H₂O formed