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

Apr 11, 2016

6.77 x ${10}^{22}$ molecules of $C {l}_{2}$.

#### Explanation:

First you need to write the correct BALANCED chemical reaction equation. Then you can determine how many moles of $C {l}_{2}$ would be produced from the available moles of Cl in 10.0g of $A l C {l}_{3}$. Finally, you will convert this value into the number of molecules using Avogadro's number.

2AlCl_3 → 2Al + 3Cl_2
So, TWO moles of $A l C {l}_{3}$ will produce THREE moles of $C {l}_{2}$.

The molecular weight of $A l C {l}_{3}$ 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 $C {l}_{2}$.

0.1125 * 6.022 x ${10}^{23}$ molecules/mole = 6.77 x ${10}^{22}$ molecules of $C {l}_{2}$.