2AlCl3 --> 2Al + 3Al2 If you produce 15 moles of Al, how many grams of AlCl3 decomposed?

1597,5 gm AlCl3 decomposed

Explanation:

2AlCl3 --> 2Al + 3Al2 If you produce 15 moles of Al, how many grams of AlCl3 decomposed?

FIRST, you need to write down the the correct reaction.  Your equation makes no sense

here is the correct question

2AlCl3 --> 2Al + 3Cl2    If you produce 15 moles of Al, how many grams of AlCl3 decomposed?

for every mole of Al produced, we decomposed a mole of AlCl3

so we produced 15 moles of Al, so we decomposed 15 moles of AlCl3

15 moles are 15 X 106.5 = 1597,5 gm AlCl3 decomposed