Chemistry help PLEASE, limiting reactions?

Question

Calculate moles of the indicated product produced when 24.0g of each reactant is used.

2Al + 3Cl2 produces 2AlCl3 find (AlCl3)

I have tried and tried on these problems. I just do not understand them. The book doesn't explain it well enough and this is an online class. Any help is really appreciated! Thanks.

Answer 1

Al has a mass of 27, and Cl2 has a mass of 71. If you use 24 g of each, you use 24/27 moles Al and 24/71 moles Cl2. You use less Cl2, so Cl2 is the limiting reagent.

To calculate AlCl3, divide the number of moles of Cl2 by 3, then multiply by 2.

48/213 = 16/71 moles AlCl3

Answer 2

I see the person above did not answer your question, nor was it very clear. I will try to be more helpful.

In limiting reactant equations, let's use an analogy of cheese sandwiches:

Start with six slices of bread and two slices of cheese. How many sandwiches can you make? Even though you have extra bread, you are "limited" to making two sandwiches by the amount of cheese on hand. Cheese is the "limiting reactant."

Now on to your specific question...

There are 24.0 grams of each reactant used. From this, we must determine how many moles of each reactant we have. Then, by looking at the balanced equation, we must convert to how many moles we can obtain of the product.

We know that there are 27.0 grams of Aluminum in one mole (from periodic chart). Start off your calculations with the "given" value of 24.0 g Al for the first value. Convert it to moles of Al, then convert to moles of AlCl3:

24.0g Al X (1 mole Al / 27.0 g Al)(2 mole AlCl3 / 2 mole Al) = 0.889 mole AlCl3

We see that for every two moles Al, there are 2 moles of the product, AlCl3. Our amount produced is 0.889 moles AlCl3.

Similarly, now calculate how much AlCl3 could be produced from the Cl2. Only here, there are two moles AlCl3 for every THREE moles Cl2:

24.0 g Cl2 X (1 mole Cl2 / 71.0 g Cl2)(2 mole AlCl3 / 3 mole Cl2) = 0.225 mole AlCl3.

Here we see that we only produce 0.225 mole AlCl3, which means this is our answer, and thus Cl2 is our limiting reactant.

Always try to set-up the equation so each corresponding value cancels out the previous value, until you get the value you are looking for (for example, here we went from grams of Cl2 to moles Cl2 to moles AlCl3).

I hope this helps!

Answer 3

The first thing you need to do is see which reacting is limiting - right?  So here we go!

There are a few ways to find the limiting reactant but the way I like best is to convert the moles of each reactant into moles of the product you are concerned with. If you use this procedure you find your limiting reactant and you determine your theoretical yield in one step using dimensional analysis.

Let's give it a try!

Quick note: remember that when you're working with reactions like this and doing stoichiometry you will be comparing amounts in moles!

For Aluminum:

24 g Al x (1 mol Al / 26.9 g Al) x (2 mol AlCl3 / 2 mol Al)

                  molecular wt of Al         ratio from the equation

When you cancel all of your units and put this in the calculator you get: 0.89 mol AlCl3

For Chlorine:

24 g Cl2 x (1 mol Cl2 / 70.9 g Cl) x (2 mol AlCl3 / 3 mol Cl2)

You get: 0.22 mol AlCl3

Therefore Chlorine is the limiting reactant (because we predict a lower yield based on how much chlorine is present) and the theoretical yield of the reaction is 0.22 mol AlCl3

There are easier ways to do this, but they aren't as reliable and won't get you the right answer every time like this method will!

Online chemistry class?  You're very brave!

Hope this helps!