Question

Chromium(III) oxide reacts with hydrogen sulfide (H2S) gas to form chromium(III) sulfide and water: Cr2O3(s) + 3H2S(g) LaTeX: \longrightarrow⟶Cr2S3(s) + 3H2O(l) To produce 560 g of Cr2S3, how many grams of H2S are required?

Answer 1

Answer:

285 g

Explanation:

Let's consider the following balanced equation.

Cr₂O₃(s) + 3 H₂S(g) ⟶ Cr₂S₃(s) + 3 H₂O(l)

The molar mass of H₂S is 34.1 g/mol and there are 3 mol × 34.1 g/mol = 102 g in the balanced equation.

The molar mass of Cr₂S₃ is 200.19 g/mol and there are 1 mol × 200.19 g/mol = 200.19 g in the balanced equation.

The mass ratio of H₂S to Cr₂S₃ is 102 g: 200.19 g. The mass of H₂S required to produce 560 g of Cr₂S₃ is:

560 g Cr₂S₃ × (102 g H₂S/200.19 g Cr₂S₃) = 285 g H₂S

Answer 2

From the stoichiometry of the reaction 268.8 g of H2S is required.

The equation of the reaction is;

Cr2O3(s) + 3H2S(g) ⟶Cr2S3(s) + 3H2O(l)

Number of moles of Cr2S3 produced =  560 g/200 g/mol = 2.8 moles

From the balanced reaction equation;

3 moles of H2S produced 1 mole of Cr2S3

x moles of H2S is required to produce 2.8 moles moles of  Cr2S3

x = 3 moles × 2.8 moles/1 mole

= 8.4 moles of H2S

Hence;

Mass of H2S required = 8.4 moles of H2S × 34 g/mol =268.8 g

Learn more about stoichiometry: https://brainly.com/question/9743981