Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s) + 3Cl2(g) -> 2AlCl3(s)

You are given 19.0g of aluminum and 24.0g of chlorine gas.

a)If you had excess chlorine, how many moles of of aluminum chloride could be produced from 19.0g of aluminum?
= 0.704mol AlCl3

b)If you had excess aluminum, how many moles of aluminum chloride could be produced from 24.0g of chlorine gas, Cl2?
=0.255mol AlCl3

By comparing your answers for Parts A and B, you can determine which reactant is limiting. Keep in mind that the limiting reactant is the one that produces the lesser amount of product.

what exactly do I do to solve for part "c"?????

c)What is the maximum mass of aluminum chloride that can be formed when reacting 19.0g of aluminum with 24.0g of chlorine?
answer:_____g AlCl3

is it?: 0.704mol AlCl3-0.255molAlCl3=0.449mol AlCl3 x 226.6g/mol = 119.7g AlCl3?

First, I don't get 0.255 for mols AlCl3 formed from 24.0 g Cl2.

I get 24/71 x (2 mols AlCl3/3 mols Cl2) = 0.338 x 2/3 = 0.225 mols AlCl3.
Check my arithmetic.

Then the answer to part c is simple. Notice that the problem says, and I quote, "Keep in mind that the limiting reactant is the one that produces the lesser amount of product." So which produces the lesser amount of product? Of course that is the 24 g Cl2 so 0.225 mols AlCl3 (which is smaller than 0.704) is the # mols and that times the molar mass AlCl3 will give you the mass AlCl3. I get 0.225 x 133.3 = 30.03 g which is rounded to 30.0 to three significant figures. Check my work closely. I think 30.0 is the answer to part c but you try it, too.

ok...so I shouldn't have multiplied through by 2. That makes a lot more sense. Thank you, yet again, and it came out alright.

(oh...slightly dislexic...yes...I meant 0.225 moles, that would have helped if I had noticed before & typed it in correctly.)

Chemlast champer vacancies

how do you do part a?

To solve for part c, you need to determine the maximum mass of aluminum chloride that can be formed when reacting 19.0g of aluminum with 24.0g of chlorine.

First, calculate the number of moles of aluminum and chlorine using their respective molar masses.

Molar mass of aluminum (Al) = 26.98 g/mol
Molar mass of chlorine (Cl2) = 70.91 g/mol (Note: We use the molar mass of Cl2 because it is a diatomic molecule)

Moles of aluminum = Mass of aluminum / Molar mass of aluminum
= 19.0 g / 26.98 g/mol
= 0.705 mol (rounded to three decimal places)

Moles of chlorine = Mass of chlorine / Molar mass of chlorine
= 24.0 g / 70.91 g/mol
= 0.338 mol (rounded to three decimal places)

Now, we have the number of moles for both reactants.

Next, we need to determine the limiting reactant, which is the one that produces the lesser amount of product. In this case, we can see that aluminum produces 0.705 mol of aluminum chloride (as calculated in part a), while chlorine produces 0.338 mol of aluminum chloride (as calculated in part b).

Since chlorine produces a lesser amount of aluminum chloride, it is the limiting reactant. Therefore, we will use the moles of chlorine to calculate the mass of aluminum chloride.

Finally, we can calculate the mass of aluminum chloride using the molar mass of aluminum chloride (AlCl3), which is 133.34 g/mol.

Mass of aluminum chloride = Moles of aluminum chloride x Molar mass of aluminum chloride
= 0.338 mol x 133.34 g/mol
= 45.02 g (rounded to three decimal places)

Therefore, the maximum mass of aluminum chloride that can be formed when reacting 19.0g of aluminum with 24.0g of chlorine is approximately 45.02 grams.

I tried this answer & it was incorrect.