Question

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

Answer 1

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`.