Question

1. -A mixture of aluminum metal and chlorine gas reacts to form aluminum chloride (AlCl3): 2Al(s) +

3Cl2(g) → 2AlCl3(s). How many moles of aluminum chloride will form when 5 moles of chlorine gas react
with excess aluminum metal?

Answer 1

3.33 moles

Step-by-step explanation:

We know that our limiting reactant is chlorine so we just have to figure out how much of the product aluminum chloride (AlCl3) would be formed from 5 moles of chlorine gas. As we can see in the balanced equation, the mole ratio of chlorine gas (Cl2) to aluminum chloride is 3:2 based on the coefficients of the species. All you have to do from here is divide the amount of moles of chlorine gas by its own coefficient and then multiply the number you receive by the coefficient of AlCl3.

5/3=1.66

1.66*2=3.33