A10.00 g sample of a soluble barium salt is treated with an excess of sodium sulfate to precipitate 11.21 g baso4 (m- 233.4). which 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)

The salt is barium chloride.

Explanation:

Moles of barium sulfate =

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

Then 0.0480 moles will be produced from:

of .

Mass of used = 10.00 g

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

Molar mass of = 208.33 g/mol

The nearest answer to our answer is .

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

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