Al2(SO4)3 +3Ca(HCO3)2==>3CaSO4 +2Al(OH)3+6CO2
Al₂³⁺(SO4)₃²⁻ + 3Ca²⁺(HCO3)⁻₂⁻==> 3Ca²⁺SO4²⁻+2Al(OH)3 +6CO2
M,sare anhidra= 342g/mol
M,cristalohidrat=666g/mol
M,Al(OH)3= 78 g/mol
din ecuatie,rezulta
1mol cristalohidrat.......1 mol sare anhidra......2molAl(OH)3
666g.......................................................................2x78g
133,2g.....................................................................m=...........calculeaza !!!!!