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:
<em>Initial mass MnO₂ = 0.6467g</em>
<em>Recovered mass = 0.3104g</em>
Mass that reacted = 0.6467g - 0.3104g = 0.3363g
<em>Moles MnO₂ -Molar mass: 86.9368g/mol-:</em>
0.3363g * (1mol / 86.9368g) = 3.868x10⁻³ moles MnO₂
<em>Moles Cl⁻:</em>
3.868x10⁻³ moles MnO₂ * (2mol Cl⁻ / 1mol MnO₂) = 7.737x10⁻³ moles Cl⁻
<em>Moles of AlCl₃ and mass -Molar mass AlCl₃: 133.34g/mol-:</em>
7.737x10⁻³ moles Cl⁻ * (1mol AlCl₃ / 3mol Cl⁻) = 2.579x10⁻³ moles AlCl₃
2.579x10⁻³ moles AlCl₃ * (133.34g / mol) =
<em>0.3439g of AlCl₃</em> are present in the sample.
The percent is:
0.3439g of AlCl₃ / 1.1701g * 100 =
<h3>29.39% of AlCl₃ in the sample</h3>
