104,4 kg x kg ykg
Cu CO3 +2 HCl =CuCl2 + CO2 + H2O
124kg 2.36,5kg 135kg
MCuCO3=64 +12 +3.16=124------->1kmol=124kg
MCuCl2=64 +3.35,5=135------> 1kmol=135kg
MHCl=1+35,5=36,5------>1mol=36,5 kg
se calculeaza puritatea CuCO3
p=mp.100: mt
mp=p.mt:100
mp=87.120:100=104 ,4 kg CuCO3 pur
se afla x
x=104,4 .2.36,5:124=61,46 kg HCl [ md ]
ms=md.100:c
ms=61,46.100:30=204,86 kg sol. HCl
se afla y
y=104,4 .135:124=113,66 kg CuCl2