Respuesta :

Answer:

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

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