Question

A salt contains only barium and one of the halide ions. A 0.158 g sample of the salt was dissolved in water, and an excess of sulfuric acid was added to form barium sulfate (BaSO4), which was filtered, dried, and weighed. Its mass was found to be 0.124 g. What is the formula of the barium halide?

Answer 1

Answer:

BaBr2

Explanation:

Halide ions are the ions of the elements of group 17 (or 7A), which have 7 electrons in their valence shell, so they need to gain one electron to be stable. If we noted as X the halide, the ion must be $X^{-}$.

Barium is a metal of group 2 (or 2A) and has 2 electrons in its valence shell, so it must lose these 2 electrons to be stable, and it will form the anion $Ba^{2+}$. These elements will form an ionic bond, with formula BaX2.

Sulfuric acid has formula H2SO4, and the reaction is:

BaX2 + H2SO4 --> BaSO4 + 2HX

So, for the stoichiometry between BaX2 and BaSO4, and denoting by y the molar mass of the halide, and the molar mass of BaSO4 = 137.33 g of Ba + 32.06 g of S + 4x16 g of O = 233.39

1 mol of BaX2 ------------- 1 mol of BaSO4

(137.33 + 2y) g of BaX2 ------------ 233.39 g of BaSO4

0.158 g of BaX2 --------------------- 0.124 g of BaSO4

By a simple direct three rule:

0.124x(137.33 + 2y) = 0.158x233.39

17.02892 + 0.248y = 36.87562

0.248y = 19.8467

y = 19.8467/0.248

y = 80.03 g/mol

By consulting the periodic table, the halide element that has a molar mass close to 80.03 g/mol is Bromide (Br), which has molar mass 79.9 g/mol. So the salt has formula BaBr2.