When 3.00 grams of the anhydrous blue colored CoCl₂ is exposed to moist air it forms 5.50 grams of the red-violet CoCl₂ ∙ xH₂O hydrate.Determine the number of moles of water that combine with one mole of CoCl₂.

Question

Quizplus · Accepted Answer

The mass of water added is 5.50 g - 3.00 g = 2.50 g. The molar mass of CoCl2 is approximately 129.83 g/mol, so 3.00 g of CoCl2 is 3.00 g / 129.83 g/mol = 0.0231 mol. The molar mass of water is 18.02 g/mol, so 2.50 g of water is 2.50 g / 18.02 g/mol = 0.1387 mol. The ratio of moles of water to moles of CoCl2 is 0.1387 mol / 0.0231 mol ≈ 6, indicating 6 moles of water per mole of CoCl2.

When 300 Grams of the Anhydrous Blue Colored CoCl2 Is Exposed

When 300 Grams of the Anhydrous Blue Colored CoCl₂ Is Exposed