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.
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.
2 answers
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
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