Aluminum reacts with Chlorine gas to form AlCl3. [2Al(s)+3Cl2(g)=2AlCl3(s)]
Given 11.0g of Al and 16.0g Cl2, if you had excess Al, how many moles of AlCl3 could be produced from 16.0g of Cl2?
You post is confusing. I can't tell if it's a limiting reagent problem or not. I'll work both.
If you had 11.0 g Al and all the Cl2 needed, you could produce how many mols of AlCl3? That's
mols Al = 11.0/27 = 0.407
mols AlCl3 = 0.407 x (2 mol AlCl3/2 mol Al) = 0.407 x 2/2 = 0.407 mols AlCl3. g = mols x molar mass.
If you had 16.0 g Cl2 and all the Al needed, how much AlCl3 could be formed? That's
mols Cl2 = 16.0/71 = 0.225 mols.
mols AlCl3 = 0.225 x (2 mols AlCl3/3 mols Cl2) = 0.225 x 2/3 = 0.15 mols AlCl3. g = mols x molar mass
The answers are above for each (11 or 16) with excess other reagent. If they are together you will be able to produce the SMALLER of the two values.
To determine the number of moles of AlCl3 that can be produced from 16.0g of Cl2, we need to use stoichiometry and convert the given mass of Cl2 to moles.
Step 1: Find the molar mass of Cl2.
The molar mass of Cl2 is calculated by adding up the atomic masses of two chlorine atoms.
Molar mass of Cl2 = 2 * atomic mass(Cl) = 2 * 35.45 g/mol = 70.90 g/mol
Step 2: Convert the mass of Cl2 to moles.
Using the molar mass of Cl2, we can calculate the number of moles of Cl2.
Number of moles = mass / molar mass
Number of moles of Cl2 = 16.0 g / 70.90 g/mol = 0.226 mol
Step 3: Use the balanced chemical equation to determine the molar ratio between Cl2 and AlCl3.
The balanced chemical equation for the reaction 2Al(s) + 3Cl2(g) → 2AlCl3(s) shows the stoichiometric ratio between Cl2 and AlCl3.
From the equation, we can see that three moles of Cl2 react to produce two moles of AlCl3. Therefore, the molar ratio of Cl2 to AlCl3 is 3:2.
Step 4: Calculate the number of moles of AlCl3.
To find the number of moles of AlCl3, we can use the molar ratio between Cl2 and AlCl3 and the number of moles of Cl2.
Number of moles of AlCl3 = (Number of moles of Cl2) * (Molar ratio of AlCl3/Cl2)
Number of moles of AlCl3 = 0.226 mol * (2/3) = 0.150 mol
Therefore, from 16.0g of Cl2, assuming excess Al, we can produce 0.150 moles of AlCl3.
To determine the number of moles of AlCl3 that can be produced from 16.0g of Cl2, you need to use the given balanced chemical equation:
2Al(s) + 3Cl2(g) → 2AlCl3(s)
First, calculate the number of moles of Cl2 using its molar mass:
Molar mass of Cl2 = 35.45 g/mol + 35.45 g/mol = 70.90 g/mol
Number of moles of Cl2 = Mass of Cl2 / Molar mass of Cl2
= 16.0g / 70.90 g/mol
= 0.226 moles of Cl2
Next, use the stoichiometric coefficients from the balanced chemical equation to determine the moles of AlCl3 produced:
From the balanced equation, the ratio between Cl2 and AlCl3 is 3:2.
Therefore, moles of AlCl3 = (0.226 moles of Cl2) × (2 moles of AlCl3 / 3 moles of Cl2)
≈ 0.1507 moles of AlCl3
Therefore, approximately 0.1507 moles of AlCl3 can be produced from 16.0g of Cl2 when excess Al is present.