1. Din Cp =>md=Cp×ms/100=8×150/100=12g
CH3-COOH
n=m/M=12/60=0.2 moli acid acetic
MCH3-COOH=12×2+4+32=60g/mol
M(CH3-COO)2Ca=2(24+3+32)+40=158g/mol
2×60g....................100g
2CH3-COOH + CaCO3 -> (CH3-COO)2Ca + CO2 + H2O
12g...................x=10g CaCO3
220g.............22.4 L CO2
22g................x=2,24 L CO2
220g................158g sare
22g.................x=15.8 g sare