5a. The balanced chemical equation for the reaction of hydrogen sulfide (H2S) and oxygen (O2) to produce water (H2O) and sulfur dioxide (SO2) is:
2 H2S + 3 O2 → 2 H2O + 2 SO2
Using the equation, we can calculate the amount of H2S needed to produce 66.6 g of H2O.
First, we need to convert the mass of H2O to moles:
66.6 g H2O × (1 mol H2O/18.02 g H2O) = 3.7 mol H2O
According to the balanced equation, 2 moles of H2S are needed to produce 2 moles of H2O. Therefore, we can set up a proportion to find the amount of H2S needed:
2 mol H2S / 2 mol H2O = x mol H2S / 3.7 mol H2O
Solving for x, we get:
x = (2 mol H2S / 2 mol H2O) × 3.7 mol H2O = 3.7 mol H2S
Finally, we can convert the moles of H2S to grams:
3.7 mol H2S × (34.08 g H2S/mol H2S) = 125.9 g H2S
Therefore, 125.9 grams of H2S are needed to produce 66.6 grams of H2O.
5b. The balanced chemical equation for the reaction of hydrogen sulfide (H2S) and chromium(III) chloride (CrCl3) to produce chromium(III) sulfide (Cr2S3) and hydrochloric acid (HCl) is:
3 H2S + 2 CrCl3 → Cr2S3 + 6 HCl
According to the stoichiometry of the balanced equation, 3 moles of H2S react with 2 moles of CrCl3 to produce 1 mole of Cr2S3.
First, we need to convert the mass of H2S to moles:
123.7 g H2S × (1 mol H2S/34.08 g H2S) = 3.63 mol H2S
Using the mole ratio from the balanced equation, we can determine the number of moles of Cr2S3 produced:
3.63 mol H2S × (1 mol Cr2S3/3 mol H2S) = 1.21 mol Cr2S3
Finally, we can convert the moles of Cr2S3 to grams:
1.21 mol Cr2S3 × (151.99 g Cr2S3/mol Cr2S3) = 184.1 g Cr2S3
Therefore, 184.1 grams of Cr2S3 are produced using 123.7 grams of H2S.