How do you solve this problem? Help! any metals react with halogens to give meta

Question

How do you solve this problem? Help! any metals react with halogens to give metal halides. For example, 2 Al (s) 3 Cl2(g) 2 AlCl3 (s) If you begin with 13.5 g of aluminum, 66.7 g of AlCl3. A) you will need 23.6 g C12 for complete reaction and will produce B) you will need 11.8 g Cl2 for complete reaction and will produce 49.0 g of AlCl3. C) you will need 53.2 g Cl2 for complete reaction and will produce 66.7 gof AlCl3. D) you will need 26.6 g Cl2 for complete reaction and will produce 49.0 g of AlCl3 E) none of the above

Explanation / Answer

2 Al (s) + 3Cl2 = 2AlCl3

or, Al + (3/2) Cl2 = AlCl3

According to the reaction 1 mole of Al reacts with 3/2 moles of Cl2 to give 1 mole of AlCl3.

Molar mass of Al = 27 g/mol

Molar mass of Cl2 = 71 g/mol

Molar mass of AlCl3 = 133.5 g/mol

13.5 g Al = 13.5/27 = 0.5 mole Al

According to the reaction 1 mole of Al reacts with 3/2 moles of Cl2 to give 1 mole of AlCl3.

So mole of 0.5 moles of Al reacts with 0.5* 3/2 or 0.75 moles of Cl2 to give 0.5 mole of AlCl3.

0.75 moles of Cl2 = 0.75 * 71 = 53.2 g Cl2

0.5 mole of AlCl3 = 0.5 * 133.5 = 66.7 g AlCl3

So answer is option C.

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