Answer:
165.726 g.
Explanation:
- For the balanced equation:
<em>Cr₂O₃ + 3H₂S → Cr₂S₃ + 3H₂O,</em>
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:
<u><em>Using cross multiplication:</em></u>
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.
