How many grams of NH4ClO4 are produced in a reaction that consumes 1.65 kg of AlCl3?

Question

Please explain, show work help
I got 3Al(s) + 3NH4ClO4(s) -> Al2O3(s) + AlCl3(s) + 3NO(g) + 6H2O(g)

So 3NH4ClO4:AlCl3

So 3:1 ratio
1650g:X
1650/3 gives you 550 grams

I got this wrong and want to know what I did wrong, thanks

Answer 1

You didn't transpose kg to grams, other then that you did it right,

Answer 2

You forgot to figure in molecular weights. Reactions happen mole to mole, not gram to gram. And it should be "grams of NH4ClO4 consumed for a reaction that produces 1.65 kg of AlCl3," based on the reaction you wrote.

1650 g AlCl3 * (1 mole/132.5 g AlCl3) * 3 moles NH4ClO4/ 1 mole AlCl3 * (117.5 g NH4ClO4/ mole) = 4390 g = 4.39 kg