2 Al + 3 Cl2 = 2 AlCl3

114g Al

186g Cl2

to use all of the Al how much Cl2 is required

Question

Chemistry

Sherlyn Gordon

24 February, 16:50

2 Al + 3 Cl2 = 2 AlCl3

114g Al

186g Cl2

to use all of the Al how much Cl2 is required

+3

Answers (2)

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Sidney · Answer 1

1 mol Al. 26.9815 g Al. *. 2 mol AlCl. 3. 2 mol Al. = 0.100 mol AlCl3. The balanced equation says: 3 mol Cl2 produces ... that the initial mass of Al was larger than required to react with all of the Cl2. The sum of the masses.

Explanation:1 mol Al. 26.9815 g Al. *. 2 mol AlCl. 3. 2 mol Al. = 0.100 mol AlCl3. The balanced equation says: 3 mol Cl2 produces ... that the initial mass of Al was larger than required to react with all of the Cl2. The sum of the masses.

2 Al + 3 Cl2 = 2 AlCl3114g Al186g Cl2to use all of the Al how much Cl2 is required

2 Al + 3 Cl2 = 2 AlCl3

114g Al

186g Cl2

to use all of the Al how much Cl2 is required