If 15.0 g of each reactant is present, what is the mass of product produced?

Question

Here is the formula (balanced): 2AI (s)+ 3Cl2 ---> AICl3

I just need help finding the product that is produced from 15 of AI and 15g of Cl2, thanks!

Answer 1

Hello, A H.,

2AI (s)+ 3Cl2 ---> AICl3

The equation isn't balanced: There needs to be one more aluminum chloride to account for the 2 Al and 6 Cl coming in.

2AI (s)+ 3Cl2 ---> 2AICl3

Convert the Al and Cl₂ masses into moles of each:

Al : 0.556 moles

Cl₂ : 0.212 moles

The molar ratio of Cl2 to Al is 1.5 Cl2 per 3 Al. We can see that there won't be nearly enough Cl2 to react with all of the aluminum. Cl₂ is the limiting reagent. We can assume all of it reacts, so use the entire 0.212 moles of Cl₂ to determine how much Al is consumed.

The molar ratio of Al to Cl₂ is (2 moles Al/3 moles Cl₂), or 2/3.

Determine the moles Al consumed:

(0.212 moles Cl₂)*(2 moles Al/3 moles Cl₂) = 0.141 moles Al consumed

That would leave 0.414 moles of Al unreacted. That is 11.192 grams.

The molar ratio of the product to the Al is 2 moles AlCl₃ per 2 moles Al, or 1 to 1.

That means we formed 0.141 moles of AlCl₃

Convert the moles AlCl₃ produced to grams: (0.141 moles AlCl₃)*(133.35 g/mole) = 18.81 grams AlCl₃.

We should be able to account for ther full 15 grams of reactants, regardless of how they are ti4ed up in molecules.

18.81 grams AlCl₃ produced

18.81 grams AlCl₃ produced

11.192 grams Al left behind

30,0 grams total

It's all accounted for.

Bob