Answer:
1597,5 gm AlCl3 decomposed
Step-by-step 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