Question

A 0.6467-g portion of manganese dioxide was added to an acidic solution in which 1.1701 g of a chloride-containing sample was dissolved. Evolution of chlorine took place as a consequence of the following reaction: After the reaction was complete, the excess was collected by filtration, washed, and weighed, and 0.3104 g was recovered. Express the results of this analysis in terms of percent aluminum chloride.

Answer 1

Answer:

29.39% of AlCl₃ in the sample

Explanation:

Based on the reaction:

MnO₂(s) + 2Cl⁻ + 4H⁺ → Mn²⁺ + Cl₂(g) + 2H₂O

We can find the amount of chloride in solution with the amount of MnO₂ that reacted as follows:

Initial mass MnO₂ = 0.6467g

Recovered mass = 0.3104g

Mass that reacted = 0.6467g - 0.3104g = 0.3363g

Moles MnO₂ -Molar mass: 86.9368g/mol-:

0.3363g * (1mol / 86.9368g) = 3.868x10⁻³ moles MnO₂

Moles Cl⁻:

3.868x10⁻³ moles MnO₂ * (2mol Cl⁻ / 1mol MnO₂) = 7.737x10⁻³ moles Cl⁻

Moles of AlCl₃ and mass -Molar mass AlCl₃: 133.34g/mol-:

7.737x10⁻³ moles Cl⁻ * (1mol AlCl₃ / 3mol Cl⁻) = 2.579x10⁻³ moles AlCl₃

2.579x10⁻³ moles AlCl₃ * (133.34g / mol) =

0.3439g of AlCl₃ are present in the sample.

The percent is:

0.3439g of AlCl₃ / 1.1701g * 100 =

29.39% of AlCl₃ in the sample