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vodomira [7]
3 years ago
8

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

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)
Chemistry
1 answer:
aliina [53]3 years ago
8 0

Answer:

The salt is barium chloride.

Explanation:

BaX_2++Na_2SO_4\rightarrow BaSO_4+2NaX

Moles of barium sulfate =\frac{11.21 g}{233.38 g/mol}=0.0480 mol

According to reaction, 1 mol of barium sulfate is produced from 1 mol of BaX_2.

Then 0.0480 moles will be produced from:

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

Mass of BaX_2 used = 10.00 g

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

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

Molar mass of BaX_2 = 208.33 g/mol

The nearest answer to our answer is BaCl_2=208.2 g/mol.

The correct answer barium chloride with molar mass of 208.2 g/mol.

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