How many molecules of chlorine gas would be produced if 10.0g of aluminum chloride decomposed into aluminum precipitate and chlorine gas?

1 Answer
Apr 11, 2016

Answer:

6.77 x #10^22# molecules of #Cl_2#.

Explanation:

First you need to write the correct BALANCED chemical reaction equation. Then you can determine how many moles of #Cl_2# would be produced from the available moles of Cl in 10.0g of #AlCl_3#. Finally, you will convert this value into the number of molecules using Avogadro's number.

#2AlCl_3 → 2Al + 3Cl_2#
So, TWO moles of #AlCl_3# will produce THREE moles of #Cl_2#.

The molecular weight of #AlCl_3# is 133.33g/mol, so 10.0g is 10.0/133.33 = 0.075mol

With our equation ratio of 3/2 this will produce 0.1125 mol #Cl_2#.

0.1125 * 6.022 x #10^23# molecules/mole = 6.77 x #10^22# molecules of #Cl_2#.