2 Al (s) + 3 Cl2 (g) è 2 AlCl3 (s) How many grams of aluminum chloride are formed when 650.02g of chlorine reacts? Whatis the classification of the reaction? Which element is being oxidized?

First, we need to determine the molar mass of each element and compound involved in the reaction.

The molar mass of Al (aluminum) is approximately 26.98 g/mol. The molar mass of Cl2 (chlorine) is approximately 70.90 g/mol. The molar mass of AlCl3 (aluminum chloride) is approximately 133.34 g/mol.

Next, we need to determine the number of moles of chlorine (Cl2) that are reacting. We do this by dividing the mass of chlorine by its molar mass.

650.02 g Cl2 * (1 mol Cl2 / 70.90 g Cl2) = 9.17 mol Cl2

According to the balanced chemical equation, 3 moles of Cl2 react to produce 2 moles of AlCl3. Therefore, we can set up a ratio to determine the number of moles of AlCl3 that are produced when 9.17 moles of Cl2 react.

(2 mol AlCl3 / 3 mol Cl2) * 9.17 mol Cl2 = 6.11 mol AlCl3

Finally, we can convert the number of moles of AlCl3 to grams by multiplying by the molar mass of AlCl3.

6.11 mol AlCl3 * (133.34 g AlCl3 / 1 mol AlCl3) = 815.04 g AlCl3

So, 815.04 grams of aluminum chloride are formed when 650.02 grams of chlorine react.

The classification of the reaction is a synthesis reaction, as it involves two or more simple substances combining to form a more complex substance.

In this reaction, aluminum is being oxidized. This is because it is losing electrons in the reaction to form Al3+ ions in AlCl3.