student was asked to determine the percentage of the components of a mixture of

Question

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 (BasO4). The mass of the sa- mple of the mixture used was 3.21 g. 5. A 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 1.43g The student treated the insoluble residue of Mg(OH)2 and BaSO4 with 3M HCI, 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

Q.5.1.(a): Given the total mass of the sample = 3.21 g

mass of KBr obtained = 1.43 g

Percent of KBr in the mixture = (1.43 g / 3.21 g) x 100 = 44.55 % (answer)

(b): mass of Mg(OH)2 obtained = 1.10 g

Hence percent of Mg(OH)2 in the mixture = (1.10 g / 3.21 g) x 100 = 34.27 % (answer)

(c): mass of BaSO4 obtained = 0.58 g

Hence percent of BaSO4 in the mixture = (0.58 g / 3.21 g) x 100 = 18.07 % (answer)

(2): Mg(OH)2 + 2HCl ------- > MgCl2 + 2H2O

(3): total mass of the substance recovered = 1.43 g + 1.10 g + 0.58 g = 3.11 g

Hence percent recovery = (3.11 g / 3.21 g) x 100 = 96.88 % (answer)

(4): Percent error = 100 - 96.88 = 3.12 % (answer)

(4)

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