5. A student was asked to determine the percentage of the components of a mixtur

Question

5. A student was asked to determine the percentage of the components of a mixture of potassium bromide (KBr), magnesium hydroxide, Mg(OH)2, and barium sulfate (BaSO.. The mass of the sa- mple of the mixture used was 3.21 g. The student extracted the KBr from the mixture with water and filtered insoluble Mg(OH)2 and BaSO, from the solution, containing the KBr. After evaporating the filtrate, the student recovered and dried the KBr and found it weighed 1.43 g. The student treated the insoluble residue of Mg(OH)2 and BaSO4 with 3M HCl, dissolving the Mg(OH)2. The student then decanted the supernatant liquid containing aqueous MgCl2 from the insoluble BaSO4. After drying the solid, the student recovered 0.58 g of BaSO4

Explanation / Answer

1.

(a)

Mass % of KBr = 1.43 * 100 / 3.21 = 44.5 %

(b)

Mass % of BaSO4 = 0.58 * 100 / 3.21 = 18.1 %

(c)

Mass % of Mg(OH)2 = 1.10 * 100 / 3.21 = 34.3 %

2.

Mg(OH)2 (s) + 2 HCl (aq.) ----------> MgCl2 (aq.) + 2 H2O (l)

3.

Total mass of substances recovered = 1.41 + 0.58 + 1.10 = 3.11 g.

% recovery = 3.11 * 100 / 3.21 = 96.9 %

4.

% error = (3.21 - 3.11) * 100 / 3.21 = 3.1 %

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