Question

630g MgCO3 puritate 80% }=> gaz +Ca(OH)2 ; oxid solid + HCl
a) R. Chimica b)ms Ca(OH)2 c=1% c) ms HCl 1200g sol HCl

Answer 1

Se calculeaza puritatea MgCO3
630.80:100=504 g MgCO3 PUR

       504 g        z         x
      MgCO3 =MgO + CO2
     84g           40        44

x=504.44:84=264 g CO2

   264g     yg
    CO2 + Ca(OH)2 =CaCO3 +H2O
     44g       74g

y=264.74:44=444 g Ca(OH)2   reprezinta md solutiei (masa subst .dizolvate)

c=md.100:ms
ms=md .100:c
c=444.100:1=44400g sol. Ca(OH)2


Se afla cantit.de MgO  obtinuta in prima ecuatie.(z)

z=504.40:84=240 g MgO

       240g    a g
       MgO +2HCl= MgCl2 +H2O
      40 g     2.36,5 g

a=240.2.36,5:40=438 g HCl   (acesta este md pentru sol. de HCl

c=md.100:ms
ms=md .100:c
ms=438.100:12=3650 g sol. HCl                          NU AI TRECUT CORECT
                                                                             CONCENTRATIA SOLUTIEI
                                                                            AM SCRIS 12  DAR NU
                                                                              STIU DACA ESTE ASA



Aici se calculeaza masele moleculare ale subst. din ecuatii


MMgCO3=24+12+3.16=84
MMgO=24 +16=40
MCO2=12+2.16=44
MCa(OH)2=40+2.16+2=74
MHCl=1+35,5=36,5