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.

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 =

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

can you attach the