When 4.00 grams of aluminum chloride reacts with excess fluorine, how many liters of chlorine gas are produced?

1.008 L

Explanation:

To solve this question you need to know the equation for the reaction. The equation for the reaction will be:

2 AlCl3 + 3 F2 → 2 AlF3 + 3 Cl2

We have excess fluorine so aluminum chloride will be the limiting reagent of the reaction. The coefficient of aluminum chloride is 2 while the coefficient of chlorine gas is 3. If the molecular mass of the aluminum chloride is 133.34 g/mol then the number of chlorine gas made in moles will be:

4g/ (133.34 g/mol) * (3/2)= 0.045 moles

Assuming the reaction happens in STP, the volume will be: 0.045 moles * 22.4 L/mol= 1.008 L