Question

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). 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)

Answer 1

The salt is barium chloride.

Step-by-step explanation:

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

Moles of barium sulfate =
`(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:

`(1)/(1)* 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.