a 10.00 gram sample of a soluble barium salt is treated with an excess of sodium sulfate to precipitate 11.21 grams BaSO4, (M= 2

Question

horsena [70]

4 years ago

6

a 10.00 gram sample of a soluble barium salt is treated with an excess of sodium sulfate to precipitate 11.21 grams BaSO4, (M= 2

33.4). which barium salt is it?

Chemistry

2 answers:

shtirl [24] · Answer 1 · 2021-12-01T07:56:27+03:00

shtirl [24]4 years ago

5 0

Hope it helps. All the best!

MrRa [10] · Answer 2 · 2021-12-01T10:20:14+03:00

For the answer to the question above, first divide the
BaSo4 and 233.4 and it will look like this.

(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt
Then,
<span>(10.0 g) / (0.0480 mol) = 208.3 g/mol </span>
<span>So the answer would be BaCl2.
</span>I hope my answer helped you.