Question

se dizolvă 0,3 moli de calciu 140 grame de apă. se cere:
a) ecuației reacției chimice
b) concentrația procentuală a soluției formate
c)numărul de moli de gaz degajat în urmă reacții
A Ca=40
AO=16
AHg=1

Answer 1

      12g  yg         zg               wg
    Ca +2H2O =Ca(OH)2 + H2                                                                                             40g  2.18g      74g           2g

1mol Ca---------40 g                                                                                                         0,3 moli Ca---------xg                                                                                                      x=0,3.40=12 g Ca
MCa(OH)2 =40+2.16 +2=74--------->1mol=74g                                                                Se calculeaza cantit.de apa consumata in reactie=y                                                      y=12.2.18:40=10,8 g H2O  consumata                                                                    140-10,8=129.2 g H2O ramasa pentru solutia ceruta                                                      Se calculeaza md solutiei=z                                                                                   z=12.74:40=22,2 g Ca(OH)2                                                                                            Se aplica formula concentratiei                                                                                    c=md.100: ms                                                                                                                  ms= md + mapa=22,2+129,2=151,4 g sol                                                   c=22,2.100:151,4=14,6 la suta
Se calc. cantit .de hidrogen din ecuatie                                                                  w=12.2:40=0,6 g H2                                                                                      1molH2--------2gH2                                                                                                      Xmoli H2 -------0,6g H2                                                                                                X=0,6:2=0,3 moli H2