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1、*高中化學(xué)方程式大全一、非金屬單質(zhì)(F2, Cl2, O2, S, N 2, P, C, Si , H)1.氧化性:F2+H=2HF (陰暗處爆炸)F2+Xe(過量)=XeE2F2(過量)+Xe=XeF (XeF 4是強(qiáng)氧化劑,能將 Mr2+氧化為MnO)nF2+2M=2MFn(函示大部分金屬)2F2+2HO=4HF+。水是還原劑)2F2+2NaOH=2NaF+OfHOF2+2NaCl=2NaF+CbF2+2NaBr=2NaF+Bi2F2+2NaI=2NaF+I27F2(過量)+l2=2IF7F2+Cl2(等體積)=2ClF (ClF 屬于類鹵素:ClF+H2O=HF+HClO )3F2(過量

2、)+Cl2=2ClF3 (ClF 3+3H2O=3HF+HClO)3Cl2+2P 1 2PCl3點(diǎn)燃3Cl2+2Fe2FeCl3Cl2+H點(diǎn)您2HCl (將H2在Cl2點(diǎn)燃;混合點(diǎn)燃、加熱、光照發(fā)生爆炸 )Cl2+PClPCl5 Cl2+2Na-2NaClCl2+CuCuCl2Cl2+2Fe2+=2Fe3+3Cl )Cl2+2FeCl2=2FeCl3 (在水溶液中:Cl2+2NaBr=2NaCl+Bi2Cl2+2Br =2Cl +Br2Cl2+2KI=2KCl+I 2Cl2+2I =2Cl +I 23Cl 2(過量)+2KI+3H2O=6HCl+KIO3Cl2+I +3H2O=6h!+6Cl

3、+IO35Cl2+12+6H2O=2HIO+10HCl5Cl2+12+6H2O=10Cl +IO3 +12H+Cl2+NaS=2NaCl+SjCl2+S2 =2Cl +SJCl2+HS=2HCl+Sj ( 水溶液中:Cl2+HaS=2hi+2Cl +SJCl2+SO+2H2O=HSQ+2HClCl2+SO+2H2O=4H+SO2 +2ClCl2+HQ=2HCl+QCl2+hbQ=2H+Cl +Q盧牌2Q+3FeFe3C4q+k=KOS+hiH2S2S+CCS>S+Zm-ZnSS+Fe = FeS (既能由單質(zhì)制取,又能由離子制?。㏒+2C一CL2S (只能由單質(zhì)制取,不能由離子制?。?S

4、+2Al=Al 2S3 (只能由單質(zhì)制取,不能由離子制取)催化劑N2+3H2 '高溫高壓、'2NHN2+3Ba-BaAbN2+6Rb-2RbNP4+6H24PHH2+2Li 2LiH2、還原性S+Q_ilLsQN2+3Mg " MgNkN2+3CaCaN2N2+6Na2Na3NN2+6K-2RNN2+2AI 二 l: 2AlNP+3Na-N&P2P+3Zn-Zn3P2S+HSQ(濃)3=3SQT +2H2QS+6HNQ 濃)一HbSQ+6NQT +2H2QS+4H(+6=6NQ T +2HQ+SCt3S+4HNQ 稀)3SQ+4NQ +2H2Q3S+4H+4

5、NQ3 3SQ+4NQ +2H2Q/電Na+Q2NQ 4P+5Q點(diǎn)燃P4Qo(常寫成P2Q)2P+3XLE=2PX (X表示 F2, CI2, Br2) PX 3+X2 =PXP4+20HNQ 濃)=4H3PQ+20NQT +4HQC+2Fr-CF4C+2CICCI4C+G(足量)堡LcQ2C+G(少量)=£E=2CQC+C(Q=2CQC+HQ=CQ+H 生成水煤氣)2C+SiQ?=Si+2CQ(制得粗硅)Si(粗)+2CI SiCI 4 (SiCI 4+2H=Si()+4HCI)Si+2NaQH+HQ=NaSiQ3+2H T(Si+2QH +HQ=SiQ; +2H4 )Si(粉)+

6、Q/ESL|SiQ2Si+CJSSLSiC(金阿J砂)(CI 2+Hag=H+CI +HCIQ)(CI 2+2QH =CI +CIQ +H2Q)(CI 2+2QH=CI +CIQ +H2Q)(3CI 2+6QH5CI +CIQ3 +3H2Q)(3S+6QH 2S2 +SQ2 +3H2Q)(4P+3QH +3HQ=PHT +3HPQ )3,歧化反應(yīng)CI2+HQ=HCI+HCIQ (加堿或光照促進(jìn)歧化:CI2+2NaQH=NaCI+NaCIQ+2QCI2+2Ca(QH)2=CaCI2+Ca(CIQ) 2+2H2Q3CI2+6KQH 濃)-5KCI+KCIQ3+3H2Q 3S+6NaQ42N&

7、;S+NaSQ+3H2。4P+3KQIO )+3HQ=PHT +3KHPQ 11P+15CuSQ+24HaQ=5CuP+6HPQ+15H2SQ 3C+Ca尸 CaC+CQT 3C+SiQ 二'SiC+2CQT、金屬單質(zhì) (Na,Mg,Al,Fe,Cu )的還原性2Na+HP_2NaH4Na+O 2=2N&O 2Na 2O+O-2N&Q2Na+O '廣 NaO 2Na+S=Na 2s (爆炸)2Na+2HIO=2NaOH+H2Na+2NH=2NaNH+Hd2Na+2HO=2Na+20T+H2 T2Na+2NH=2Na+2NH2 +H2 T4Na+TiCl U4Na

8、Cl+TiMg+C2MgCb Mg+B4MgBis2Mg+O ' 2MgOMg+SMgS 2Cu+S=Cu2s (Cu 2s只能由單質(zhì)制備)Mg+2HOMg(OH>+H42Mg+TiCL燎融Ti+2MgCl2Mg+2RbC戶匚MgCb+2Rb2Mg+CO 2MgO+CMg+HS=MgS+H Mg+HSO=MgSOH T2Al+3Cl 2: ' 2AlCl 32Mg+SiO2MgO+Si_+2+(Mg+2H=Mg +H4 )4Al+3O2=2Al2Q (常溫生成致密氧化膜而鈍化,在氧氣中燃燒)4Al(Hg)+3O 2+2xHO=2(Al 2Q.xHzO)+4Hg (鋁汞齊)

9、高溫4Al+3MnQ2Al 2Q+3Mn_ _高溫 _ _ 上一一2Al+Cr 2O=Al 2O+2Cr (鋁熱反應(yīng))2Al+Fe2Q, Al ?Q+2Fe2Al+3FeO ' Al 2Q+3Fe2Al+6H+=2Al3+3H2 T2Al+6H+=2Al "+3H42Al+2NaOH+2HO=2NaAlO+3H T2Al+2OH +2HbO=2AlQ +3H T2Al+6HCl=2AlCl 3+3H2 T2Al+3H2SO=Al 2(SO4) 3+3H2 T 2Al+6H2sO(濃)=Al 2(SO4) 3+3SO+6H2O (Al,Fe 在冷,濃的 HSQ,HNQ 中鈍化)

10、Al+4HNO(稀)=Al(NO3)3+NO +2H2O Al+4H+NO =Al3+NOT +2H2O_ _ _占懶_ _向痼 _ _2Fe+3Br2=2FeBr33Fe+2O 二旦Fe3。2Fe+O=2FeO (煉鋼過程)Fe+I LFeI 2Fe+S=FeS (FeS既能由單質(zhì)制備,又能由離子制備)3Fe+4HO(g) Fe3Q+4H2 TFe+2HCl=FeCl2+H2 TFe+2l4=Fe2+H TFe+CuCl2=FeCl2+CuFe+Cif=Fe2+Cu JFe+SnCl4=FeCl2+SnCl2(鐵在酸性環(huán)境下,不能把四氯化錫完全還原為單質(zhì)錫Fe+SnCl2=FeCl2+Sn

11、JFe+Sn+=Fe+Sn三.非金屬氫化物(HF,HCl,H 2O,H2S,NH3)金屬氫化物(NaH)1、還原性:4HCl(濃)+MnO=MnC2+C12 T +2H2O4H+2Cl +MnO且=MK+Cl2 T +2HO4HCl(濃)+PbO=PbCl2+C12 T +2H2O4H+2C1 +PbO=Pt2+C12 T +2H2O4HCl(g)+O 22Cl 2+2H2O16HCl+2KMnO=2KCl+2MnCL+5Cl2 T +8H2O16 H+10Cl-+2MnO =2Mr2+5Cl2 T +8H2O6HCl+KClQ=KCl+3Cl 2 T +3H2O6H+5Cl +ClO3 =3

12、Cl 2 T +3H2O14HCl+&Cr2Q=2KCl+2CrCl 3+3Cl 2 T +7HO14H+6Cl +Cr2。2 =2Cr3+5Cl2 T +7H2O2H2O+2F=4HF+O2HCl+F2=2HF+Cl2 (F 2氣與 HCl、HBr、HI、H2S、NH3氣體不能共存)2HBr+Cl2=2HCl+Br2 (Cl 2氣與 HBr、Hk H2S NH氣體不能共存)2H2S+3O(足量)型L2SQ+2H2。2 H2S+O(少量)彗!=2S J +2H2O2HS+SO=3SJ +2H2OH2S+HSO(濃)=SJ +SQT +2H2O3H2S+2HNO ?。?3SJ +2N0T

13、 +4hbO3H2S+2I4+2NO =3S J +2N0T +4H2O5Hs+2KMnG3HSO=2MnS©K2SO+5s J +8H2O5HS+2MnO +6H+=2Mr2+5S J +8H2O3H2S+KCr2O7+4H2SO=Cr2(SO4) 3+K2SO+3S J +7HO3H2S+Cr2。2 +8H+=2Cr3+3Sj +7HOHS+4NaQ+2H2O=NaSO+6NaOHHaS+4Ns2O+2HaO=8N,a+SO 2 +OH加熱2NH+3CuO3Cu+N+3HaO2NH+3Cl 2=N2+6HCl8NH+3Cl 2=N2+6NHClNH+NaN(2>HCl=Na

14、Cl+N2T +2H2ONH3+NO +/=N4 +2H2O“占昌催化劑4NH+3c2(純氧)W2N2+6H2O4NH+5QA 4NO+6HO4NH+6NO=5沖6H2O (用氨清除 NO)NaH+HO=NaOH+K (生氫齊)NaH+HO=Na+OH +Hd加熱4NaH+TiCl 4Ti+4NaCl+2H 2 fCaH+2H2O=Ca(OH) J +2Hd2、酸性:4HF+SiQ=SiF4+2hbO (可測定礦樣或鋼樣中SQ的含量,玻璃雕刻)4HF+Si=SiF4+2H2 T2HF+CaCb=CaE+2HClhbS+Fe=FeSj +H2 THaS+CuCb=CuS +2HCl (弱酸制強(qiáng)酸

15、的典型反應(yīng) )HS+Cif+=CuSj +2H+HS+2AgN0=A&Sj +2HNOH2S+2Ag+=A1SJ +2HHS+HgCl2=HgS; +2HC1HS+Hj=HgSj +2H+HS+Pb(NG) 2=PbSj +2HNO(鉛試紙檢驗(yàn)空氣中 H2S)h2S+Pt2+=PbSj +2H+H2S+2Ag=AgS+H仙銀器在空氣中變黑的原因 )2NH(液)+2Na=2NaNb+HJ (NaNH 2+H2O=NaOH+NH )3 . NH3的堿性:NH+HX=N4X (X : F、Cl、Br、I、S)NH+HNO=NHNONH+H+=NH+2NH+HaSO=(NH4) 2SONH+H

16、+=NH+NH+NaCl+HO+C®=NaHCONHCl (侯德榜制堿:用于工業(yè)制備小蘇打,蘇打)_ _+_-NH+HS=NHHSNH+hbS=NH +HS4 .不穩(wěn)定性:高溫高溫扃溫2HFH2+F22HClhb+Cl22H2O2H2+Q2HO=2HO+O加熱H2Shb+S催化劑2NH: N2+3H2HI-Ha+l2四.非金屬氧化物(SC3、SO、N2ONON2Q、NO、N2Q、N2。、CO CO、SiO2、F2Q、心。、Cl2O ClzQ、ClzQ、Cl 2。、ClO2)1、低價(jià)態(tài)的還原性:(SO2、CO NO)2SO+Q+2HO=2HSO (這是SO在大氣中緩慢發(fā)生的環(huán)境化學(xué)反應(yīng)

17、)催化劑2so+Q2SQ so+n-noSO+C12+2HO=HSC+2HC1SO+Br2+2H2O=HSO+2HBrSO+l2+2H2O=HSQ+2HICl 2+SQ+2H2O=4H+SO2 +2ClBr2+SO+2H2O=4H+SO2 +2BrI 2+SO+2H2O=4Hi+SO2 +2I2NO+O=2NONO+NQ2NaOH=2NaNa用于制硝酸工業(yè)中吸收尾氣中的NO+NQ2OH=2NONO和 NO)2CO+O ' 2CO加德CO+CuO Cu+CO加熱3CO+FQ2Fe+3CO2、氧化性:SO+2HS=3S+2HO加熱|SO+2KIK2SO+12催化劑CO+HO=5L NO+2

18、KI+H2O=NO+2+2KOH(不能用淀粉KI溶液鑒另U澳蒸氣和 NO) 4NO+H2S=4NO+SOH2O加熱2NO+Cu 4CuO+N加熱N2O+ZnZnO+N芭甥 I CO+2M"=2MgO+C (CQ不能用于撲火由 Mg,Ca,Ba,Na,K等燃燒的火災(zāi))SiO2+2H2Si+2H2OSiO2+2Mg2MgO+Si3、與水的作用:SQ+H>O=HSOSO+HO=HSQ2 -SO+HO=2H+SO23NQ+HO=2HN9NO (NQ不是硝酸的酸酊)NaQ+HO=2HNQN2Q+hbO=2hi+2NOBQ+HO(冷水)=2HP。P2Q+3HO(熱水)=2H3PQ (P2O

19、5極易吸水,可作氣體干燥劑)BQ+3I4SO(濃)=2H3PO+3S。CO+HO=HCOCl2O+HO=2HClOCl2O+HO=2HClOCl2O+HbO=2H+2clOJ4.與堿性物質(zhì)的作用:SO+2NH+HO=(NH0 2SOSO+(NH4) 2SO+HO=2N4HSO2NHHSO+HaSO=(NH4)2SO+2H2O+2SOT (硫酸工業(yè)尾氣處理)SO+Ca(OH)2=CaSOj +H2O (不能用澄清石灰水鑒別 SO和CO.可用品紅鑒別)SO+MgO=MgSOSO+Ca(OH)2=CaSOj +H2OCO+NH+HO=NHHCOCO+2NH(過量)+HO=(NH) 2CO(NH4)2

20、COL (NH)2CO+2HO催3fIICO+2NH曜口(NH2)2CO+HO (工業(yè)制取尿素)CO+2NaOH±量)=NaCO+HO CO(過量)+NaOH=NaHCO CO+Ca(OH)2(過量)=CaOI4O 2CO(過量)+Ca(OH)2=Ca(HC(3) 2 CO+CaCGHO=Ca(HCO 22OH+CO=CGf +HOOH+CO=HCOCa+2OH +CO=CaCOj +HOO1 +CO=HCOCO+CaCG+HO=Ca+2HCOCO(不足)+2NaAlO2+3H2O=2Al(OH)3 J +NaCOCG+3H2O+AQ =Al(OH) 3 J +CO2CO(足)+Na

21、AlO2+2H2O=Al(OH)3 J +NaHCOCO+2H2O+AlQ =Al(OH) 3 J +HCOCO+GH5ONa+HO=CHOHj +NaHCOCG+GHCT+HO=CH5OH; +HCOSiO2+Ca道且CaSQ (煉鋼造渣)SiO2+2NaOH=N2SiO3+HO(常溫下強(qiáng)堿緩慢腐蝕玻璃 )SiO2+Na2cos1=N&SiO3+CO (制取玻璃)SiO2+CaCQ3iLCaSiO3+CQ (制取玻璃) 2NO+2NaOH=NaNeNaNHO2NG+2OH=NO +NO +HONO+NQ2NaOH=2NaNHO (制取硝酸工業(yè)尾氣吸收 )NO+NQ2OHT=2NO +

22、H2O五.金屬氧化物1 .低價(jià)態(tài)的還原性:6FeO+O=2F&QFeO+4HN=Fe(NQ) 3+NQ+2H2OFeO+4H+NO =Fe"+NOT +2H2O2 .氧化性:NaO+2Na2Na2O (此反應(yīng)用于制備 NaQMgO Al 2Q幾乎沒有氧化性,很難被還原為Mg, Al. 一般通過電解制 Mg和Al.Fe2O+3H=2Fe+3HO(制還原鐵粉)Fe3Q+4H2-3Fe+4H2OCuO+H /Cu+HO2Fe3Q+16HI =6FeI 2+8H2O+2I22Fe3Q+16H+4I =6Fe2+8H2O+2l2FeO+Fe j£iL3FeO (煉鋼過程中加入

23、廢鋼作氧化劑 )FeO+CJLFe+CO (高溫?zé)掍撜{(diào)節(jié) C含量)2FeO+S亶國L2Fe+SiO2 (高溫?zé)掍撜{(diào)節(jié)Si含量)3 .與水的作用:NaO+HO=2NaOHNaO+HO=2Na+20H2N&Q+2H2O=4NaOHS2NaQ+2H2O=4Na+40H +Q T(此反應(yīng)分兩步:Na2Q+2HaO=2NaOH+6;2H2Q=2HO+O HaQ的制 備可利用 類似的 反 應(yīng):BaO2+HaSQ(稀尸=BaSQ+HO)MgO+bO=Mg(OH(緩慢反應(yīng))4 .與酸性物質(zhì)的作用:NaO+SON&SON&O+CO=NaCOMgO+S3=MgSONaO+2HCl=2NaC

24、l+HONaO+2H=2Na+H2O2NaQ+2CO=2NaCO+O TNa>O+H>SO(冷,稀)=NaSQ+HOMgO+bSQ=MgSO>H2OMgO+2譯Md+hbOAl2O+3HSO=Al 2(SO4) 3+3H2OAl 2O3+6H+=2Al "+3H2OAl2O+2NaOH=2NaAlQHO(Al 2Q兩性氧化物)Al 2Q+2O1=2AlO2 +HaOFeO+2HCl=FeCb+HzOFeO+2H=Fe2+HOFe2O+6HCl=2FeCl3+3H2OFe2Q+6H+=2Fe3+3HzOFeQ+8HCl=FeCl2+2FeCl3+4HO Fe3Q+8

25、H+=2Fe3+Fe2+4H2O六.含氧酸1、氧化性:4HC1Q+3H2s=3HSQ+4HClC1O3 +3H2s=6l4+SO2 +ClHClO3+HI=HIO3+HClClO3 +I =IO3 +Cl3HClO+HI=HIQ+3HCl3HClO+I-=IO3 +3H+ClHClO+HSO=H2SO+HClHClO+HSQ=3H+SO2 +ClHClO+HQ=HCl+hbO+OT HClO+HQ=+Cl +HO+OT(氧化性:HClO>HClQ>HClQ>HClQ,但濃,熱的HClQ氧化性很強(qiáng))2H2SO(濃)+C=COT +2SOT +2H2O2H2SO(濃)+S=3SO

26、T +2H2OHSO+Fe(Al)室溫下鈍化6H2s0(濃)+2Fe=Fe2(SO4)3+3SO T +6H2O2H2SO(濃)+Cu=CuS0+SQT +2HOHSO(濃)+2HBr=SOT +Br2+2hbO(不能用濃硫酸與 NaBr制取 HBr)HSO(濃)+2HI=SOT +I2+2HO(不能用濃硫酸與 NaI制取HI)HSO(?。?Fe=FeS(4>Ha T2H+Fe=Fe2+hb THSO+2HS=3SJ +3H2O4HN0 濃)+G=COT +4NOT +2HO6HN 濃)+S=H2SO+6N0T +2H2O5HN 濃)+P=HPQ+5NOT +HO5HN0 稀)+3P+2

27、H出 3H3PO+5NO5H+5NO-+3P+2HO3H3PO+5NOT6HN0 濃足)+Fe=Fe(NO) 3+3NOT +3H2O4HNQ濃)+Fe(足)=Fe(NO) 2+NOT +2HO(先得 Fe3+,在 Fe過量時(shí)再生成 Fe2+的鹽)4HN0 稀足)+Fe=Fe(NO)3+NOT +2H2O4H+NO+Fe=Fe3+NOT +2H2O30HNO+8Fe=8Fe(NQ) 3+3N2OT +15屋。30 H+6NO +8Fe=8Fe3+3NaOT +15HO36HNOH0Fe=10Fe(NQ)3+3N4 +18H2O36H+6NO +10Fe=8Fe3+3lM +18HO30HN08

28、Fe=8Fe(NQ) 3+3NHNO+9H2O30 H+3NO +8Fe=8Fe3+3NH+9HO4Zn+10HN1稀)=4Zn(NO3) 2+N2OT +5H2O +2+4Zn+10H+2NO =4Zn +N2OT +5H2O4Zn+10HNQ ?。?4Zn(NO3) 2+NHNO+3H2。4Zn+10H+NO =4Zn2+NH+5H2。2、還原性:HSG+X+HO=HSO+2HX (X 表示 Cl2,Br 2,I 2)HSQ+X2+H2O=4H+SQ2-+X 2H2SQ+Q=2HbSO+2-2H2SO+Q=4H+SQH2SO+HQ=H2SQ+HOH2SQ+HO=2H+SO2 +HbO5l4

29、SO+2KMnO=2MnS©K2SQ+2h2SO+3H2O 5H2SO+2MnO =2Mr2+4H+3SO2 +3H2OH2SO+2FeCl3+H2O=HSO+2FeCl 2+2HCI HSQ+2Fe3+HO=4H+2Fe2+ +SO423.酸性:(不揮發(fā)性酸制取揮發(fā)性酸)(不揮發(fā)性酸制取揮發(fā)性酸)(不揮發(fā)性酸制取揮發(fā)性酸)(不揮發(fā)性酸制取揮發(fā)性酸)(強(qiáng)酸制弱酸酸)(工業(yè)制磷肥)H2SO(濃)+CaE =CaSO+2HFTH2SO(濃)+NaC0NaHSGHCl TH2SO(濃)+2NaCN*SO+2HCl TH2SO(濃)+NaNO=NaHSGHNOT 3HsO(濃)+Ca(PO4

30、) ,3CaSO+2H3PO 2H2SO(濃)+Ca3(PO4) 42CaSO+Ca(H2PO) 2 3HNO+Ag3PO=HPQ+3AgNO3H+Ag3PO=HPO+3Ag+ 2HNG+CaCO=Ca(NO) 2+H2O+COT2H+CaCO=Ca2+HO+COT(用HNO和濃HSQ不能制備H2S, HI, HBr, SO2等還原性氣體)4HPO+Cd(PO4) 2 £3Ca(H2PO)2(重鈣)HPO(濃)+NaBr 4 NaHzPO+HBrT (不揮發(fā)性酸制取揮發(fā)性酸 ,磷酸是非氧化性酸)HPO (濃)+Na0NaHPO+HI T4.不穩(wěn)定性: 2HCl屋"2HCl+

31、O4 (保存在棕色瓶中)4HNOMNO T +O T +2H2O (保存在棕色瓶中)HSO=HO+SOT (在加熱或酸性條件下分解 )HCO-HO+COT (在加熱或酸性條件下分解 ) TSiO,-H2SiO3+HOH2SiO3 SiO 2 J +H2OH2S2O=H2O+S +S0仙在加熱或酸性條件下分解 ) 七.堿1 .低價(jià)態(tài)的還原性:4Fe(OHh+Q+2HO=4Fe(OH)2 .與酸性物質(zhì)的作用:2NaOH+SO量)=NaSO+HOOH +SO=SO2 +HONaOH+SO足)=NaHSOOH+SO(足)=HSO2NaOH+SiO=NaSiO3+H2。OH+SiO2=SiO32 +HO

32、2NaOH+AlQ=2NaAlQ+HO 2OH+Al2Q=2AlO2 +HO2KOH+C2=KCl+KClO+HO Cl2+2OH=Cl +ClO +H2ONaOH+HCl=NaCl+2OHI+OH =HbONaOH+2S(足)=NaHS+2OOH+HaS=HS+HaO2NaOH+2S(少量)=NaS+2Ho 2OH+HS=S" +2HO3NaOH+AlC3=Al(OH) 3 J +3NaCl3OH +Al 3+=Al(OH) 3 JNaOH+Al(OH)=NaAlQ+2HO(AlCl 3和 Al(OH) 3 哪個(gè)酸性強(qiáng)?)OH+Al(OH) 3=AlO2 +2H2OCa(OH)2+

33、2NHC出2CaCl2+2NHT +2HO (實(shí)驗(yàn)室制 NH)NaOH+NCl 區(qū) NaCl+NHT +HbOMg(OH)+2NHCl=MgCl2+2NH H2O (Al(OH) 3+NHCl 不溶解)Ba(OH)2+HbSQ=BaSO J +2H2O2H+2OH +Ba2+SO2 =BaSOj 2H2。3 .不穩(wěn)定性:Mg(OH)MgO+HD 2Fe(OH)3Fe2O+3H2O 八、鹽1.氧化性:(在水溶液中) 2FeCl3+Fe=3FeCl22Al(OH) -Al 2c3+3HOCu(OH)JCuO+HO2AgOH=AgD+HO2Fe3+Fe=3Fe2+2Fe3+Cu=2Fe2+Cu +2

34、FeCl3+Cu=2FeCl2+CuCl2 (用于雕刻銅線路版)2Fe3+Zn=2Fe2+Zn2+2FeCl3+Zn(少量)=2FeCl2+ZnCl2FeCl3+Ag=FeCl2+AgCl JFe(SQ) 3+2Ag=FeSO+Ag2SQ J 2FeCl 3+H2s=2FeCL+2HCl+S J 2FeCl3+2KI=2FeCl 2+2KCl+I 22Fe3+Cl-+2Ag=2Fe2+2AgCl J(較難反應(yīng))Fe(NO 3) 3+Ag不反應(yīng)2Fe3+HS=2Fe2+2H+Sj2Fe3+2I =2Fe2+12FeCl 2+Mg=Fe+MgGlFe2+Mg=Fe+M&NaN®N

35、HCl=NaCl+N2 T +2H2O (實(shí)驗(yàn)室制氮?dú)?NH+NO- =Na f +2H2O2.還原性:2FeCl2+3Cl 2=2FeCl3 (在水溶液中不需加熱 ) 2Fe2+3Cl2=2Fe3+6Cl3N&S+8HNQ 稀)=6NaNO+2NOT +3S+4HO3S2-+8H+2NO-=2NOT +3S+4HO3N0SO+2HNO;稀)=3Na2SO+2NOT +F2O3SO2-+2H+2NO-=3SO2-+2NOT +HO2NaSO+Q=2NaSO (Na 2SO在空氣中易變質(zhì)) NaSO+S二 NaS2QS2-+Cl 2=2Cl-+SjNaS+Cl2=2NaCl+Sj (在水

36、溶液中)3.與堿性物質(zhì)的作用:Ca(OH)2+CuSO=Cu(OH)2 J +CaSOj (波爾多液)MgCb+2NH - H2O=Mg(OH)J +2NHClMg+2NH - hbO=Mg(OH2 J +2NH+AlCl 3+3NH H2O=Al(OH)3 J +3NHClAl 3+3NH - H2O=Al(OH)2j +3NH+FeCl3+3NH - H2O=Fe(OH3 J +3NHClFe3+3NH - H2O=Fe(OH)4 +3NH+CuSG+2NH - H2O 不足)=Cu(OHJ +(NT)2SOCif+2NH - hbO=Cu(OH)4 +2NH+Cu(OH)2+4NH3 H

37、O=Cu(NH)4(OH)2+4H2OCu(OH)2+4NH HO=Cu(NH)42+2OH+4HO 銅氨溶液CuSG+4NH- H2O 足)=Cu(NH)4SO+4HO總方程式Cif+4NH H2O=Cu(NH3)4 2+4H2。銅氨溶液AgNO+NH - HO=AgOH +NHNO2AgOH=Agp灰黑色)+HOAg2O+4NH - H2O=2Ag(NHs) 2 +2OH+3H2O 銀氨溶液AgNG+2NH - HaO=Ag(Nh) 2NO+2Ho總方程式Ag+2NH3 - HO=Ag(NH) 2+2H2。ZnSO+2NH - HaQ 不足)=Zn(OH)2 J +(NT)2SOZn2+2

38、NH - H2O=Zn(OH)2 J +2NH+Zn(OH)2+4NH3 - HO=Zn(NH) 4(OH)2+4H2。ZnSO+4NH- H2O 足)=Zn(NH) 4SO+4HOZn2+4NH - H2O=Zn(NH3) 4 2+4H2??偡匠淌交虿粨]發(fā)性酸制揮發(fā)性酸PO3-+2H=HPO-HP(4f+M=HPO-h2PQ-+H=hbPOco2-+i4=hcohco+h+=cot +H2O4 .與酸性物質(zhì)的作用:強(qiáng)酸制弱酸, NaPO+2HCl=NaHPO+2NaClNaHPO+HCl=NaHPO+NaClNaHPQ+HCl=HPO+NaClNaCG+HCl=NaHC0NaClNaHC0H

39、Cl=NaCl+H2O+COT3NaCO+2AlCl 3+3H2O=2Al(OH)3 J +3COT +6NaCl (物質(zhì)之間的雙水解反應(yīng) ) 3CO2-+2Al 3+3H2O=2Al(OH)3 J +3COT3NaCO+2FeCl3+3HO=2Fe(OH)4+3CQ+6NaCl (物質(zhì)之間的雙水解反應(yīng) )3CO2-+2Fe3+3H2O=2Fe(OH)U +3COT(物質(zhì)之間的雙水解反應(yīng))(物質(zhì)之間的雙水解反應(yīng))(物質(zhì)之間的雙水解反應(yīng))(物質(zhì)之間的雙水解反應(yīng))3NaHC吩AlCl 3=Al(OH) 3 +3COT3HCO+Al 3+=2Al(OH) 3 J +3CO T3NaHC吩FeCl3=

40、Fe(OH)3 J +3COT3HCO+Fe3+=2Fe(OH)3 J +3CO T3NaS+Al 2(SO4) 3+6HO=2Al(OH)3 J +3HST3S2-+2Al 3+3H2O=2Al(OH) 3 J +3HST3NaAlQ+AlCl 3+6HaO=4Al(OH)3 J +3NaCl 3AlO2-+Al 3+6H2O=4Al(OH)3 J3NaAlQ+FeCl3+6HO=3Al(OH)3 J +Fe(OH)3 J +3NaCl 3AlO2-+Fe3+6H2O=3Al(OH) 3 J +Fe(OH)3 JNaAlQ+NHCl+2H2O=Al(OH)3 J +NH - HbO+NaCl

41、 _ _ -+AO2+NH+2H2O=Al(OH)3j +NH - H2ONaCO+HO+C2=2NaHCOCO2-+H2O+CO=2HCONaCO+HO+2SO=2NaHSGCOT (1:2)CO2-+HO+2SO=2HSQ +CO T2NaCO(足)+HbO+SO=NaSO+2NaHCO(CO2 中的 SO不能用 N*CO洗氣)2 CO2+HaO+S0=SO2-+2HCO (2:1)NaCOSO=Na2SO+CO(1:1)co2-+so=so2-+coNaHC0SO=NaHSaCO (CO 2 中的 SO可能用 NaHC3M氣) 2HCO+SO=2HS+CO2NaHCaSO=NaSO+2C

42、O+HO2HCO+SO=SO2+2CO+hbONaSiO3+2HCl=HSiO3 J +NaCl 或 NaSiO3+2HCl+HO=HSiO4 J +2NaClSiO32-+2H+=HSiO3 J或 SiO32-+2H+HO=HSiO4 JNaSiO3+CO+2H2O=HSiO3 J +NkCOSiO32+CO+2H2O=HSiO4j +CO2-5 .鹽與鹽復(fù)分解反應(yīng)NaSO+BaCl2=BaSOj+2NaCl(沉淀不溶于鹽酸、石肖酸 )SO2-+Ba2+=BaSOj,沉淀不消失)NaSO+BaCl2=BaSOj+2NaCl (沉淀溶于鹽酸,在硝酸中生成新的沉淀SO2-+Ba2+=BaSOjN

43、aCG+BaCl2=BaCOj+2NaCl(沉淀溶于鹽酸、沉淀消失 )CO2-+Ba2+=BaCOjNaCO+CaCl2=CaCOJ +2NaCl (NaHCO3不反應(yīng)) CO2-+CsT=CaCOjAgNO+NaCl=AgCl J +NaNOAg+Cl-=AgCl JAgNO+NaBr=AgBr J +NaNOAg+Br-=AgBr JAgNO+KI=AgCl J +KNOAg+I -=AgI J3AgNO+Na3PQ=AgPO J +3NaNO 3Ag+PO3-=Ag3PO J CuSO+Na2S=CuS +NaSOCiT+S2-=CuSjFeCl 3+3KSCN=Fe(SCN»

44、3KClFe3+3SCN=Fe(SCN)3 (血紅色,用于 Fe3+的特性檢驗(yàn))6 .不穩(wěn)定性:NaSO+HSO=NaSQ+Sj +SOT +H2O &Q2-+2l4=Sj +SOT +H2ONHCl=NHI T +HCl TNHI 旦 NHT+HIT 2HI=H+I2NHI=NhlT +H2T +I2TNHHCO=NH T +H2O+COT2KNO-2KNO+QT2Cu(NQ) 32CuO+4NOf +O2 T光照2AgNO=2Ag+2NOT+QT (保存在棕色瓶中)5NHNOUMN T +2HNO9K010NHNO=8N2 T +4N0 +-2Q + +20H2OT (硝酸錢爆炸反

45、應(yīng))ZKMnA&Mn0+MnO+Q T2KC1O32KC1+3O2 T2NaHCAN&CO+H2O+COTCa(HCQ jCaCO+HO+COTCaCL CaO+COTMgCb; MgO+CO九、電離方程式1.酸的電離(H2SQ、HN HC1、HBr、HI、H3PO、HR H2SQ、CHCOOH H2CO、H2s HN QH50H HCN HC1O)HSO=2l4+SO2-或:H2SQ+2HO=2HC+SO2-HN(3=H+NQ-或:HNG+HO=HC+N (以下雷同)HC1=H+C1HBr=H+Br +HI=H+IHPOH+HPQHaPO4H+HP。HP。H+pO4HH+FH

46、aSOH+HSQ HSO 3H+So2CHCOOhf+CHCOOHCOH+ HCO3HCO 3 H+ C03+HH+HShs =h + s2HN(0=H+N02GHOXH+GHO (苯酚不是酸,顯酸性)HCN=H+CN一一.+HC1Q=H+C10HO=H+OH2h2O=H3Q)+OH2.堿的電離(NaOH KOH Ba(0H)2、Mg(OH、A1(0H) 3、NH H2O)NaOH=N+OHKOH=%OHBa(OH)2=Bs2+2OH_.2+_Mg(OH>=Mg +20HAl(OH) 3;-Al 3+3OH 酸式電離:Al(OH) 3-H+AIO 2+件ONH - H2= NH 4 +2

47、OHCa(OH)2=Cs2+2OH (澄清石灰水)Ca(OH)2=Ca2+2OH (石灰懸濁液)3.鹽的電離(NaCl、NaSQ、NaHSO NaSO、NaHSO MgSQ CaSQ Al 2(SO4)3、CuSQ AlCl 3、 AgN CHCOONa NHNO、FeCl3、Na2CO、NaHCO NaS、NaHS NaHPQ、NaHPQ NaPQ、 KI、NaBr、NaClO AgCl、CaCONaCl=Na+ClNaSQ=2Na+SO4NaHS©=H+Na+SO4NaSO=2Na+SO4NaHS©=Na+HSO (錯(cuò)誤書寫:NaHSO=Nc+H+SQ2-)MgSO=M

48、j+SO43+Al2(SQ)3=2Al +3SO4CuSO=Cif+so4AlCl 3=Al 3+3ClAgNO=Ag+NOCHCOONa=CCQO +Na+NHNQ=NH+NO-FeCl3=Fe3+3ClNaCO=2Na+ CO2+NaHC3=Na + HCO3 (錯(cuò)誤書寫:NaHC®=Na+H + CO;)NaS=2Na+ S2 +NaHS=Ns+HS (錯(cuò)誤書寫:NaHS=Ns+H+S2 )NaHPQ=Nsf+H2PqNaHPO=2Na+HP0 (錯(cuò)誤書寫:Na2HP(4=2Nci+K+PC3 )+ _ _3NaPQ=3Na+PCKI=K+INaBr=Nc+BrNaClO=N&

49、lt;a+ClOAgCl=Ag+Cl (難溶、微溶物質(zhì)在水中發(fā)生微弱電離)CaCO=Ca2+CO2 (錯(cuò)誤書寫:CaCO=C+CC2 )CaSOCaT+sO (錯(cuò)誤書寫:CaS(4=Ca2+SO2 )3、熔融電離NaCl,Na+ClMgCl2-zLMc2+2ClNa2OOL=2Na+C2Al 2OS112Al 3+3C2十、水解反應(yīng)1 .單水解-可逆水解NF4Cl+H2Q=Nhl - H2O+HCI Nhl+H2te= hf+Nhl - H2OFeCl 3+3-0= Fe(OH)3+3HCl Fe3+3H2O= Fe(OH)3+3H+AlCl 3+3H2q=ai(qh)3+3HClAl 3+3

50、H2O= Al(OH) 3+3H+CuSO+2H2X=Cu(OH)2+HSQ (金屬活動(dòng)順序表中 Mj以后的陽離子均水解 )NaHC3+H2O=H2CO+NaOH (NaHSM水解,NaHSO電離大于水解)NaCQ+HQ NaHCQ+NaOH C(C2-+H2= HCO+OHNaHC0H2(片=HCO+NaOH第一步遠(yuǎn)遠(yuǎn)大于第二步,二步不能疊加)NaSQ+HaQNaHSONaO牌02+屋0= HSQ-+OHNaHSQ+KQhkSO+NaOH第一步遠(yuǎn)遠(yuǎn)大于第二步,二步不能疊加)HSO+HO HSO+OH_ _ 2- _ -Na>S+H=NaHS+NaOHS +HbyHS+OHNaHS+bO

51、= H2S+NaOH第一步遠(yuǎn)遠(yuǎn)大于第二步,二步不能疊加)HS+H20= H2S+OHNaPO+H2O=NaHPO+NaOHPO'+HbO HP02-+0H_ _2-_-_ -NaHPQ+hb(NaHPO+NaOH HPO+噸0= HbPO+0HNaHPQ+HHPQ+NaOHH2P0-+Hb= H3P0+0HCHCOONa+Q= CHCOOH+NaOHCHCOC>HO= CHCOOH+O HGH5ONa+HQH5OH+NaOHGHO+HbO GHOH+OH2 .雙水解CHCOONHHO=CHCOOH+NH HONHF+HgHF+NH- H2OAl2S+6HO=Al(OH)3 J+H

52、ST (隔絕空氣,密封保存)MgN+6H0=Mg(0H)J +NHT (隔絕空氣,密封保存 )NaP+3HO=3NaOH+PH (隔絕空氣,密封保存 )Zn3F2+6HO=Zn(OH)4+PH4 (Zn3P2一種老鼠藥, PH劇毒神經(jīng)毒劑)CaC+2H2O=Ca(OH)4 +QH4 (隔絕空氣,密封保存 )GH5ONa+HC=GH5OH+NaOH十一、電解及電極方程式1 .電解質(zhì)溶液在惰性電極條件下,或陰極是較活潑金屬電極,陽極是惰性電極條件下的電 解申解2NaCl+2H202NaOH+HT +C12 T(-)2H2O+2e-=Hz T +20H 或 2H +2e-=Hz T(+)2Cl-2e-=Cl 2 T電解I2Cl -+2HQ20H+H T +Cl2 TCuCl2電解Cu+C12 T(-)Cu(+)2C12+2e-=Cu J-2e-=C1 2 TCu 2+2C1-= : CuJ +C12T 2CuSO2H0 - 2Cu J +Q T +2HSO(-)2Cu2+ + 4e-=2Cu J(+)2H2O - 4e-=0 2T +4H+ 或:

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