md=ms.c/100=400 .40/100=160 g Ca(OH)2 2moli Ca(OH)2=74.2=148 g deci avem un exces de 12 g
CO2+Ca(OH)2=CaCO3+H2O
2moli x
1mol.......................1moli x=2 moli CaCO3 rezultati
CaCO3+2HCl=CaCl2+CO2+H2O
1mol.......2.36,5g.........1mol
1mol.............x.................y
x=73 g de HCl consumat in reactie (md pentru această solutie)
ms=md.100/c=73.100/36,5=200g
y=1 mol CO2 Degajat in a doua reactie