Question

Chromium(III) oxide reacts with hydrogen sulphide gas to form chromium(III) sulphide and water. How many grams of hydrogen sulphide are required to produce 324.8 g of chromium(III) sulphide?

Answer 1

165.726 g.

Step-by-step explanation:

• For the balanced equation:

Cr₂O₃ + 3H₂S → Cr₂S₃ + 3H₂O,

It is clear that 1 mol of Cr₂O₃ and 3 mol of H₂S to produce 1 mol of Cr₂S₃ and 3 mol of H₂O.

• Firstly, we need to calculate the no. of moles of 324.8 g of chromium(III) sulphide:

no. of moles of Cr₂S₃ = mass/molar mass = (324.8 g)/(200.19 g/mol) = 1.62 mol.

• Now, we can find the "no. of grams" of H₂S are needed:

Using cross multiplication:

3 mol of H₂S produces → 1 mol of Cr₂S₃, from stichiometry.

??? mol of H₂S produces → 1.62 mol of Cr₂S₃.

∴ The no. of moles of H₂S are needed = (3 mol)(1.62 mol)/(1 mol) = 4.86 mol.

∴ The "no. of grams" of H₂S are needed = (no. of moles of H₂S)(molar mass of H₂S) = (4.86 mol)(34.1 g/mol) = 165.726 g.