Question

Hydrogen sulfide (H_2S)reacts with oxygen (O_2) to produce sulfur dioxide (SO_2) and water. How many grams of H_2O will be produced if 100 g of H_2S react with 100 g of O_2?

Answer 1

To find the amount of water produced, the balanced chemical equation is used. Oxygen is identified as the limiting reactant, yielding 37.5 grams of H₂O from the reaction of 100 grams each of H₂S and O₂.

Step-by-step explanation:

To calculate the amount of water (H2O) produced from the reaction of hydrogen sulfide (H2S) with oxygen (O2), we need to refer to the balanced chemical equation:

2H2S(g) + 3O2 (g) → 2H2O(l) + 2SO2(g)

First, we calculate the molar mass of H2S (34.1 g/mol) and O2 (32.0 g/mol) to find the number of moles of each reactant:

• 100 g H2S ÷ 34.1 g/mol = 2.93 moles of H2S
• 100 g O2 ÷ 32.0 g/mol = 3.125 moles of O2

Since oxygen is the limiting reactant (ratio 3:2 in the equation), we proceed with its moles:

• (3.125 moles O2) x (2 moles H2O/3 moles O2) = 2.083 moles H2O

The molar mass of H2O is 18.0 g/mol. Now we convert the moles of H2O to grams:

• 2.083 moles H2O x 18.0 g/mol = 37.5 grams of H2O

Therefore, 37.5 grams of water will be produced when 100 g of H2S react with 100 g of O2.