Question

2. 10.00 grams of a sample of hydrated PtCl4 are heated and lose 3.00 g of water. How many moles of water are combined with each mole of PtCl4?

Answer 1

Answer:

8 mol

Explanation:

Step 1: Calculate the mass of PtCl₄ in the sample

10.00 grams of a sample of hydrated PtCl₄ are heated and lose 3.00 g of water. The mass of PtCl₄ is:

mPtCl₄ = 10.00 g - 3.00 g = 7.00 g

Step 2: Calculate the moles corresponding to 7.00 g of PtCl₄ and 3.00 g of H₂O

The molar mass of PtCl₄ is 336.9 g/mol.

7.00 g × 1 mol/336.9 g = 0.0208 mol

The molar mass of H₂O is 18.02 g/mol.

3.00 g × 1 mol/18.02 g = 0.166 mol

The molar ratio of H₂O to PtCl₄ is:

0.166 mol H₂O/0.0208 mol PtCl₄ ≈ 8 mol H₂O/ 1 mol PtCl₄