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A 0.940g sample of barium chloride dehydrate is heated and 0.800g of anhydrous residue remains after cooling.how many moles of water is there

User Chicowitz
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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.

User Adin
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