Hi, I posted my question earlier but I was having some difficulty with it.

2Al+ 3Cl2-->2AlCl3
You are given 31.0g of aluminum and 36.0g of chlorine gas.

1.If you had excess chlorine, how many moles of aluminum chloride could be produced from 31.0g of aluminum?

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This is my work so far:

Cl2-36.0g/70.9= 0.507mol
Al-31.0g/26.98=1.15mol

Therefore Chlorine is the limiting reagent, since I have less that 3/2.

Cl2:AlCl3= 3:2
0.507mol (3/2)=0.76 mol AlCl3.

Please check my answer and significant figures. I am pretty sure I messed up!

Yes, you did.

The first part is not a limiting reagent since you have all of the Cl2 needed.
moles Al = 31.0/26.98 = 1.149
moles AlCl3 = 1.149 moles Al x (2 moles AlCl3/2 moles Al) = 1.149 x (1/1) = 1.149 moles AlCl3 which rounds to 1.15 to three s.f.

#2a You can do it two ways. The way I suggested, since you had to do #1 anyway, is to keep #1 in mind and work #1 with 36.0 g Cl2 and all the Al needed.
moles Cl2 in 36.0 g is
36.0/70.9 = 0.50776 (which I would have rounded to 0.508 BUT I usually carry extra places(in the calculator) and round at the end of the problem.
moles AlCl3 = 0.50776 mols Cl2 x (2 moles AlCl3/3 moles Cl2) = 0.3385 moles which rounds to 0.338 moles AlCl3. (Your error is in the conversion factor.) Then I compare moles from #1 where we had all of the Cl2 needed (1.15) with #2 where we had all of the Al we needed (0.338) and the correct value is the smaller of the two; that is 0.338 moles AlCl3 formed.
2b. The other way Bob Pursley showed you is the way you did it to determine Cl2 was the limiting reagent.
31.0/26.98 = 1.15 moles Al and
(31.0/26.98) x (3 moles Cl2/2 moles Al) = 1.72 moles Cl2 needed and we don't have that much.
(36.0/70.9 = 0.508 moles Cl2.
(36.0/70.9) x (2 moles Al/3 moles Cl2) = 0.338 moles Al needed and we have that much; therefore, your work is ok in determining the limiting reagent.

You gotta take the bleeble blabble and add it to sazzle shnazzle.

You must find the value for x and then twirk on a poster of Miley Cyrus for an hour.

To find the moles of aluminum chloride that can be produced from 31.0g of aluminum, you first need to calculate the moles of aluminum.

Given:
Mass of aluminum (Al) = 31.0g
Molar mass of aluminum (Al) = 26.98 g/mol

To calculate the moles of aluminum:

moles of Al = mass of Al / molar mass of Al
moles of Al = 31.0g / 26.98 g/mol
moles of Al = 1.15 mol

So, you calculated the moles of aluminum correctly as 1.15 mol.

Next, let's find the moles of chlorine gas (Cl2):

Given:
Mass of chlorine gas (Cl2) = 36.0g
Molar mass of chlorine gas (Cl2) = 70.9 g/mol

moles of Cl2 = mass of Cl2 / molar mass of Cl2
moles of Cl2 = 36.0g / 70.9 g/mol
moles of Cl2 = 0.507 mol

From the balanced chemical equation, we know that the stoichiometric ratio of Cl2 to AlCl3 is 3:2. This means that for every 3 moles of Cl2, we can form 2 moles of AlCl3.

Now, we need to determine which reactant is the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

To find the limiting reagent, we compare the moles of both reactants with their respective stoichiometric ratios.

For Cl2: AlCl3, the stoichiometric ratio is 3:2.
We have 0.507 mol of Cl2.

moles of AlCl3 = (moles of Cl2) * (2 moles of AlCl3 / 3 moles of Cl2)
moles of AlCl3 = 0.507 mol * (2/3)
moles of AlCl3 = 0.338 mol

According to our calculations, we have 0.338 mol of AlCl3 that can be produced. However, we need to consider the number of significant figures.

The initial mass values of both reactants have three significant figures (31.0g and 36.0g). In calculations involving multiplication or division, the final answer should be reported with the same number of significant figures as the least precise measurement involved. In this case, the least precise measurement is 36.0g of Cl2.

Therefore, the final answer should be rounded to three significant figures.

So, the correct answer is:

If you had excess chlorine, you could produce 0.338 mol of aluminum chloride from 31.0g of aluminum.