Final answer:
After heating, the mass of water lost from the hydrated barium chloride was 0.140 g, which is equivalent to approximately 0.0078 moles of water.
Step-by-step explanation:
The question asks how many moles of water were lost when a hydrated sample of barium chloride was heated to leave an anhydrous residue. Since the initial mass of the hydrated compound was 0.940 g and the remaining anhydrous residue weighed 0.800 g, we can subtract the mass of the residue from the initial mass to find the mass of the water lost:
Mass of water = 0.940 g - 0.800 g = 0.140 g
Next, we can calculate the moles of water by using the molar mass of water, which is approximately 18.015 g/mol:
Moles of water = mass of water / molar mass of water
Moles of water = 0.140 g / 18.015 g/mol
Moles of water ≈ 0.0078 mol
Therefore, the sample contained approximately 0.0078 moles of water.