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Sagot :
312.17 g iodine would react with 20.4 g aluminum completely to form aluminum iodide.
2Al + 3I2 → 2AII3
Mass of aluminum = 21.3 g
Moles of Aluminum = 21.4/ 27g/mol = 0.79mol
The stoichiometric mole ratio of iodine to Al : 3mol I2/2molAl
Moles of Iodine that would react with 0.756 mol Al =
0.79 × 3mol I2/2molAl = 1.23molI2
Mass of iodine = 1.23 molI2 * 253.81gI2/1mol I2 = 312.17g I2
Therefore, 312.17 g iodine would react with 20.4 g aluminum completely to form aluminum iodide.
An amalgam of aluminium and iodine, aluminium iodide is a chemical. The name is invariably used to describe a product with the chemical formula AlI 3, which is produced by the interaction between aluminium and iodine[4] or by the action of HI on aluminium metal. Aluminum hydroxide, hydrogen iodide, or hydroiodic acid react with metallic aluminum to produce the hexahydrate. AlI is related to the compounds chloride and bromide.
The Lewis acid 3, which is potent, will take water from the air. It serves as a reagent in the scission of particular types of C-O and N-O bonds. Additionally, it deoxygenates epoxides and cleaves aryl ethers.
Learn more about aluminum iodide here:
https://brainly.com/question/14963619
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