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 $Cl_2$.
