Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.