A salt contains only barium and one of the halide ions.?

Question

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

Answer 1

The reaction is:

BaX2 + H2SO4 ----->BaSO4 + 2HX

being X the halide ion.

1) We must know the molecular masses of the known compounds:

M (H2SO4) = 98 g/mol
M (BaSO4) = 233.3 g/mol

2) Let´s see the possible barium halides with their corresponding molecular masses:

M(BaF2) = 175.3 g/mol
M(BaCl2) = 208.3 g/mol
M(BaBr2) = 297.3 g/mol
M(BaI2)= 391.3 g/mol

3) Let´s see how many moles of the BaSO4 were produced according to reaction above:

n (BaSO4) = 0.124 g/ 233.3 g/mol = 0.00053 moles

Because we know that an excess of sulphuric acid reacted with the halide salt (it means that all the halide salt reacted), then stoichiometry of the reaction above indicate that:

BaX2 ............ +..... H2SO4 ------------>BaSO4 + 2HX
0.00053 moles.....0.00053 moles........0.00053 moles

4) All we have to do is to prove each molecular mass of the possible halides to catch the right one:

0.00053 x 175.3 g/mol (BaF3) = 0.0929 g does not coincide with the data of 0.158 g

0.00053 x 208.3 g/mol (BaCl2) = 0.110 g does not coincide with the data of 0.158 g

0.00053 x 297.3 g/mol (BaBr2) = 0.157 g that is almost the same amount 0.158 g. We have found it !!

The formula of this barium halide is:

BaBr2 (Barium bromide).

Good luck!

Answer 2

Halide Ions

Answer 3

by an extremely speedy and nasty technique in my head, SO4 = ~ninety 4 g/mol. iodide = ~127 g/mol, this is heavy sufficient to account for the preliminary pattern being heavier than BaSO4 by speedy assessment. Astatide is yet another conceivable halogen (heavier than iodide) = ~210 g/mol and this would be too extensive of a weight difference so it can not be conceivable. save in mind that Ba is two+, so which you want 2 halogen ions. ideal way is to transform 0.176 g BaSO4 (~233 g/mol) into moles (~7.55e-4 moles). Then multiply by molar mass of a BaX2, the place X = any halogen. It in basic terms comes close if X = iodide. BaI2 (~391 g/mol) and it will provide extra or less your unique 0.296 g of pattern. So that's BaI2

Answer 4

Use the solubility equilibrium 1.08E-10 = [Ba^2+][SO4^2-]