n HCOOH = 6 mol
Al(OH)₃ + 3HCOOH ----> (HCOO)₃Al + 3H₂O
3 mole HCOOH reagują z 1 molem Al(OH)₃
czyli
6 moli HCOOH przereguje z 2 molami Al(OH)₃
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n HCOOH = 6 mol
Al(OH)₃ + 3HCOOH ----> (HCOO)₃Al + 3H₂O
3 mole HCOOH reagują z 1 molem Al(OH)₃
czyli
6 moli HCOOH przereguje z 2 molami Al(OH)₃