Posted by simon on Wednesday, January 9, 2013 at 9:24am.
first, note that there are 2 Al on each side. So, we need to balance the
CO3 and the O3CO2
as written, we have
3C and 9O on the left, and
1C and 5O on the right.
But, 3O are bound to the balanced Al, leaving
1C and 3O to play with. Looks like we just need to triple the CO2:
Al2(CO3)3 --> Al2O3 + 3CO2
Left: 3C,9O
Right: 3C,9O
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