Question

o cantitate de 4,9g oxid de potasiu reacționează cu 46,6g apă. ce concentrație procentuală va avea soluția hidroxid de potasiu formată?

Answer 1

      4,9g     xg          yg
        2K + 2H2O =2 KOH + H2                                                                                               2.39   2.18g      2.56g
MH2O=2+16=18-------->1mol=18g                                                                            MKOH=39+16+1=56------>1mol=56g                                                                                x=4,9.2.18 :2.39=2,26 g H2O                                                                                         46,6-2,26=44,34 g apa ramasa pentru solutie                                                                 Se calc. md solutiei =y=cantit. de KOH                                                          Y=4,9.2.56:2.39=7,035 g KOH  =  md                                                                        c=md.100:ms                                                                                                                   ms=md +mapa=7,035+44,34=51,375  g sol.                                                                                                               c=7,035.100:51,375=13,69 la suta