A 10.00 g sample of a soluble barium salt is treated with an excess of sodium sulfate to precipitate 11.21 g BaSO4 (M- 233.4). W

Question

11 days ago

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hich barium salt is it? 2. (A) BaCl2 (M- 208.2) (B) Ba(O2CH)2 (M- 227.3) (C) Ba(NO3)2 (M 261.3) (D) BaBr (M- 297.1)

1 answer:

eduard [2.4K] · Answer 1 · 2026-03-24T17:47:15+03:00

Answer:

The salt identified is barium chloride.

Explanation:

$BaX_2++Na_2SO_4\rightarrow BaSO_4+2NaX$

The moles of barium sulfate produced are $\frac{11.21 g}{233.38 g/mol}=0.0480 mol$

per the reaction, 1 mole of barium sulfate arises from 1 mole of $BaX_2$ .

Therefore, 0.0480 moles result from:

$\frac{1}{1}\times 0.0480 mol=0.0480 mol$ of $BaX_2$ .

The quantity of $BaX_2$ used amounts to 10.00 g

Moles of $BaX_2$ = \frac{10.00 g}{\text{Molar mass}}[/tex]

$0.0480 mol=\frac{10.00}{\text{Molar mass}}$

The molar mass of $BaX_2$ is 208.33 g/mol

The closest answer to our calculation is $BaCl_2=208.2 g/mol$ .

The correct identification is barium chloride, which has a molar mass of 208.2 g/mol.

s < p < d < f