分析酸式鹽與堿反應(yīng)的兩種方法

1、分析酸式鹽與堿反響的兩種方法:?5?:?Cu-2e-=Cu2+,?2H+2e-=H2?N?O?m?:Cu+2H2O?Cu(OH)2+H2,?amolH2,?2amo,l?,?)?(C)?e?o?,?,?N?O?m?2H2O?O2+2H2,?amolH2,?amo,l?(D)?(?a?),?-?C?H?,?,?)?,?)?,?,?;?,?,?,?o?:?:?:?B?C?(226200)?e?m?a?N?a?,?,?e?u?,?m?a?,?)?,?)?a?,?(?o?5?,Ca(HCO3)2?aNaOH?,?)?(?,?5?o?5?e?O?,?a?e?m?a?,?:?e?m?a?5?,?(?,?,

2、?m?,?;?,?e?)?(?,?,?)?(?)?Ca(HCO3)2?aNaOH?Ca(HCO3)2?aNaOH?e?:(O3)=+O3O3+O3NaOH=Na+OH-?H?,?a?5?:HCO-3+OH-=CO2-3+H2O,CO2-3+Ca2+=CaCO3?,1molCa(HCO3)2?2molHCO-3,?,?NaOH?,1mol?,2molHCO-3?,?NaOH?2mol?,HCO-3?,Ca(HCO3)2?aNaOH?,11?,?,12?,?N?O?m?e?:Ca(HCO3)2+NaOH(?l?)=CaCO3+NaHCO3+H2OCa(HCO3)2+2NaOH(?)=CaCO3+N

3、a2CO3+2H2O?(?e?e?,NaHCO3?aCa(OH)2?,11?N21?:NaHCO3+Ca(OH)2(?)=CaCO3+NaOH+H2O2NaHCO3+Ca(OH)2(?l?)=CaCO3+Na2CO3+2H2O?e?Ca(HCO3)2?aCa(OH)2?2?e?:(O3)=+O3(O)=+O63?e?N?(?)CaHC2Ca22HC-HC-HC2-CaHC2Ca22HC-CaH2Ca22H-,?a?5?:HCO-3+OH-=CO2-3+H2O,CO2-3+Ca2+=CaCO3?0,1molCa(HCO3)2?2molHCO-3,1molCa(OH)2?2molOH-,2molH

4、CO-3?a2molOH-?,?5?o2molCO2-3,?+2molCa2+?5?o2molCaCO3?,Ca(HCO3)2?aCa(OH)2?,11?,?:Ca(HCO3)2+Ca(OH)2=2CaCO3+2H2O?H?e?,?l?)?O?,?m?a?,?,?:1Ca(HCO3)2+NaOH?:?)?,?,11?,?5?o?m?;?,?,12?,?5?o?,:Ca(HCO3)2?NCa(HSO3)2,Ba(HCO3)2,Ba(HSO3)2+NaOH(?NKOH)2NaHCO3+Ca(OH)2?:?,?,11?,?5?o?;?)?,?,21?,?5?o?,:NaHCO3(?NKHCO3,Na

5、HSO3,KHSO3)+Ca(OH)2?NBa(OH)23Ca(HCO3)2+Ca(OH)2?:?2?,11?,:Ca(HCO3)2?NCa(HSO3)2,Ba(HCO3)2,Ba(HSO3)2+Ca(OH)2?NBa(OH)2?,?e?m?a?,?5?,?,?,?l?5?o?:?0?:?l?:?)?Ca(HCO3)2?aNaOH?2?N?O?m,?,11?N?O?m:Ca(HCO3)2+NaOH=CaCO3+NaHCO3+H2O?(O3)?NO?,?O?a?,?5?o?Ca(HCO3)2?,?Ca(HCO3)2?a?5?o?CaCO3?:NaHCO3?;?NaOH?,?,?NaOH?aNaH

6、CO3?5?o,?NaOH+NaHCO3=Na2CO3+H2O,?e?H?m?a?m?+?Ca(HCO3)2+2NaOH=CaCO3+Na2CO3+2H2O?(?,12?N?O?m?(?,?O?Ca(HCO3)2?aNaOH?,11?,12?,12?N?O?m:Ca(HCO3)2+2NaOH=CaCO3+Na2CO3+2H2O,?(?Ca(HCO3)2?NaOH?,?O?a?,?5?o?,?Ca(HCO3)2?,?Ca(HCO3)2?a?5?o?Na2CO3?,?Ca(HCO3)2+Na2CO3=CaCO3+2NaHCO3,?e?m?a?H?m?+?Ca(HCO3)2+NaOH=CaCO3+NaHCO3+H2O,?(?,11?N?O?m?,Ca(HCO3)2?aCa(OH)2?,?N?O?mCa(HCO3)2+Ca(OH)2=2CaCO3+2H2O,?,?Ca(HCO3)2?NCa(OH)2?,?a?,?Ca(HCO3)2?aCa(OH)2?,11?a?,?,?e?,?0?5?o?a?H?)?e?e?5?,?e?:?