26.98 grams of Al and 35.45 grams of Cl2 are reacted according to the equation
2Al + 3Cl2 → 2AlCl3
How many grams of AlCl3 will be formed assuming 100% yield?
it says the answer is supposed to be 26.98g but cannot seem to reach that. I just want to know how to get there.
no the answer isnt 26.98.. its 44.45ish
first do the aluminum:
26.98 x (1 mol Al/26.98 g Al) x (2 mol AlCl3/2 mol Al) x (133.5g AlCl3/ 1 mole AlCl3) = 133.5
then the chlorine gas:
35.45 x (1 mol Cl2/70.9g Cl2) x (2 mol AlCl3/3 mol Cl2) x (133.5/1 mol) = 44.45
Do you recognize that this is a limiting reagent problem? That may be the problem. If you had posted what you did I could have pointed out the exact place you went wrong.
Determine the limiting reagent.
Using that substance, convert mols of the LR to mols of the product
Then convert mols product to grams. g = mols x molar mass.
I don't think the answer is 26.98 but more like 44.45 give or take a little. I just estimated quickly. Cl2 is the limiting reagent.
To determine the amount of AlCl3 formed, we need to use the concept of stoichiometry.
1. Start by converting the mass of Al and Cl2 to moles. To do this, divide the given masses by their respective molar masses.
- Molar mass of Al = 26.98 g/mol
- Molar mass of Cl2 = 35.45 g/mol
Moles of Al = 26.98 g / 26.98 g/mol = 1.00 mol
Moles of Cl2 = 35.45 g / 35.45 g/mol = 1.00 mol
2. Now, we can determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed. To find the limiting reactant, we need to compare the mole ratio of Al to Cl2 using the balanced equation.
From the equation 2Al + 3Cl2 → 2AlCl3, we see that the mole ratio of Al to Cl2 is 2:3. This means that for every 2 moles of Al, we need 3 moles of Cl2.
Since we have 1.00 mol of Al and 1.00 mol of Cl2, the ratio is equal. Therefore, neither reactant is in excess, and both will completely react.
3. Now that we know both reactants will be completely consumed, we can determine the amount of AlCl3 formed using the mole ratio from the balanced equation. The mole ratio tells us that for every 2 moles of Al, we form 2 moles of AlCl3.
Since we have 1.00 mol of Al, this means that 1.00 mol of AlCl3 will be formed.
4. Finally, convert the moles of AlCl3 to grams by multiplying by the molar mass of AlCl3.
- Molar mass of AlCl3 = 133.34 g/mol
Mass of AlCl3 = 1.00 mol × 133.34 g/mol ≈ 133.34 g
Therefore, assuming 100% yield, you should get approximately 133.34 grams of AlCl3.
The answer of 26.98 g given might be a typo or a mistake because the calculated amount is actually 133.34 g.