Silver nitrate is reacted with magnesium bromide and two products are formed.
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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 .
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 . 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 <em>y</em> 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.