Question

19. The formula of a hydrated salt of barium is BaCl2 ⋅ xH2O. If 1.936 g of this compound gives 1.846 g
of anhydrous BaSO4 upon treatment with H2SO4, calculate x.

Answer 1

Question #41592 – Chemistry – Other

Question:

the formula of a hydrated salt of barium is BaCl₂ · xH₂O. If 1.936 g of this compound gives 1.846 g of anhydrous BaSO₄ upon treatment with H₂SO₄, calculate x.

Answer:

General reaction equation:

$$\mathrm{BaCl_2 \cdot xH_2O + H_2SO_4 = BaSO_4 + xH_2O + 2HCl}$$

Barium chloride reacts with sulfuric acid to produce HCl and barium sulphate (BaSO₄):

y...1.864

BaCl₂ + H₂SO₄ → BaSO₄ + 2 HCl

1 mole...1 mole

208g...233g

y is the mass of "anhydrous" BaCl₂ in the 1.936 g of hydrate.

y = (208 * 1.864) / 233

y = 387.712 / 233

y = 1.664 g

One mole of BaCl₂ is 208g, so 1.664g will have 1.664 / 208 = 0.008 moles BaCl₂.

The water in the original BaCl₂·xH₂O is: 1.936g – 1.664g = 0.272g. In moles, this is 0.272 / 18 = 0.0151 moles of water.

We have 0.008 moles BaCl₂ and 0.0151 moles of water together.

x = 0.0151 / 0.008 = 1.88 ≈ 2

So, the formula of compound is BaCl₂·2H₂O

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