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

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.