Question

8 moli de hidroxid de aluminiu din ce cantitate de oxid de aluminiu se obtine ... Va rog frumos :)))​

Answer 1

Răspuns:

MAl(OH)3=27+3×16+3×1=78g/mol

MAl2O3=2×27+3×16=102g/mol

m=78×8=624g

102g 2×78g

Al2O3 + 3 H2O => 2 Al(OH)3

xg 624g

x=624×102/2×78=408g Al2O3