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

how do you do "b"

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

I kept trying it, via comparing it to a, & finally did come up with 0.255mol AlCl3...I had just gotten the question incorrect,somehow, 4x...so I had to ask. Thank you for your patience with me. I will be looking into getting a tutor shortly.

To solve part (b) of the problem, we will use the given mass of chlorine gas to determine the number of moles of aluminum chloride that can be produced.

First, we need to calculate the moles of chlorine gas (Cl2) from the given mass of 24.0g. To do this, we need to divide the mass by the molar mass of chlorine gas.

The molar mass of chlorine gas (Cl2) is calculated by adding the molar masses of two chlorine atoms:

Molar mass of Cl2 = (2 × atomic mass of chlorine)

The atomic mass of chlorine (Cl) is given as 35.45 g/mol. Therefore, the molar mass of Cl2 is:

Molar mass of Cl2 = 2 × 35.45 g/mol = 70.90 g/mol

Now we can calculate the number of moles of chlorine gas (Cl2):

Number of moles of Cl2 = Mass of Cl2 / Molar mass of Cl2 = 24.0g / 70.90 g/mol

Calculate the value:

Number of moles of Cl2 = 0.338 mol

According to the balanced equation, the stoichiometric ratio between Cl2 and AlCl3 is 3:2. Therefore, for every 3 moles of Cl2, we get 2 moles of AlCl3.

Using this ratio, we can now calculate the moles of aluminum chloride (AlCl3) that can be produced from the 0.338 moles of chlorine gas (Cl2):

Number of moles of AlCl3 = (2/3) × Number of moles of Cl2

Calculate the value:

Number of moles of AlCl3 = (2/3) × 0.338 = 0.225 mol

Therefore, if you had excess aluminum, you could produce 0.225 mol of aluminum chloride (AlCl3) from 24.0g of chlorine gas (Cl2).

To find the number of moles of aluminum chloride that could be produced from 24.0g of chlorine gas (Cl2), you need to use the given equation's stoichiometry and the molar mass of Cl2.

First, calculate the number of moles of Cl2 by dividing the given mass by its molar mass:
Molar mass of Cl2 = 2 * atomic mass of Cl
= 2 * 35.45 g/mol = 70.90 g/mol

Number of moles of Cl2 = Mass of Cl2 / Molar mass of Cl2
= 24.0 g / 70.90 g/mol
= 0.338 mol

Next, use the equation's stoichiometry to determine the ratio between Cl2 and AlCl3.
From the balanced equation: 2Al(s) + 3Cl2(g) -> 2AlCl3(s)
The stoichiometric coefficient of Cl2 is 3, meaning that it takes 3 moles of Cl2 to produce 2 moles of AlCl3.

Since the stoichiometry ratio is 3:2, divide the moles of Cl2 by its coefficient (3) and multiply by the coefficient of AlCl3 (2):
Number of moles of AlCl3 = (Number of moles of Cl2 / Coefficient of Cl2) * Coefficient of AlCl3
= (0.338 mol / 3) * 2
= 0.225 mol * 2
= 0.45 mol

Therefore, when you have excess aluminum, you could produce 0.45 moles of aluminum chloride from 24.0g of chlorine gas (Cl2).

I don't understand how you can do a without knowing how to do b.

mols Cl2 gas = 24.0/molar mass Cl2 = ??

mols AlCl3 = mols Cl2 x (2 mols AlCl3/3 mols Cl2) =
??mols Cl2 x (2/3) = xx mols AlCl3.
I hope this helps