高中無機化學方程式大全

1、高中化學方程式大全-非金屬單質(F2 ,。2, 02, s, N2, B G Si, H )1.氧化性：F2+H2= = =2HF (陰暗處爆炸)F2+Xc(i±®=XcF22F2fi±S)+Xc=XcF4 (XuF,是強氧化劑，能將氧化為MnOL)nF2+2M=2MFn(M表示大部分金屬)2F2+2Hq=4HF+O2 (水是還原劑)2F2+2NaC)H=2NaF+()F>+H2C)F2+2NaCl=2NaF+Cl2F2+2NaBr=2NaF+Br>F2+2NaI=2NaF+L7F 乂過量)+L=21 艮與+CM等體積)=2C1F (C1F 屬于類鹵素

2、：C1F+H2O=HF+HCK)3F(過量)+。2=2。巳(C1F3+3H2O=3HF+HCK)3)Cb+H_4量2Ho (將H?在Cb點燃；混合點燃、加熱、光照發(fā)生爆炸)占甦AA3Cl2+2P=2PChC12+PCl=PCh CL+2Na=2NaQ占妙占供3C12+2Fc2FcC15C12+Cu=CuCbC12+2FcC12=2FcC15 (在水溶液中：Cl2+2Fc2+=2Fc+3Cr)Cl2+2NaBr=2NaCl+Br2Clz+ZB/ =2CT +Br2C12+2KI=2KCl+ba2+2l'=2Cl'+I23cH 過重)+2KI+3H2O=6HC1+KK)33cl2+

3、I - +3H2O=6H+6C1 - +IO35Cl2+I>+6H2O=2HI()5+10HCl5Cl2+I2+6H2O=10a +IO3 -+12H+CL+Na2s=2NaCl+S Ch+S?- =2CT +S |Cb+H2s=2HC1+S 1(水溶液中：Ch+H2s=2H+2C1-+S |Cb+SO2+2H2O=H2so4+2H。C12+SO2+2H2O=4H*+SC>42- +2C1 -C12+H2()2=2HCl+O2Cl2+H2O2=2ir+Cl - +O2占供2()2+3F=Fc3()45+K=K()2S+H f=H2S2S+C=CS>S+Zn=ZnSS+Fc 4

4、 FuS （既能由單質制取，又能由高于制取）S+2Cu4Cu2s （只能由單質制取，不能由高子制?。?s+2A14Ag （只能由單質制取，不能由高子制?。┐呋瘎a3代手而T2NH3占燃W+3M 乒Mlg»2N2+3Ca=CaAN2N2+3Ba=Ba5N2N2+6Na=2Na3NN2+6K=2KNN2+6Rb=2Rb5NN+2A 畫也2A1NP4+6H =4PH3 P+3N；F=NaP2P+3Zn=Zn,P2H2+2Lj=2LiH2、還原性占供S+（）S（） 一MS+H2so* 濃 T=3S（）21 +2比。S+6HN（）3（濃：=H2so4+6NO2 1 +2H2（）S+4H+6=

5、6NC>21 +2H2O+SO； 3s+4HN（%（?。﹒=3S（）2+4NO f +2H2（）3S+4H*+4 NO ； =3SO2+4NO t +2H2O放電N2+（）2=2N（）占燃4P+5（）9PQE常寫成 P（）5）PX3+X,=PX52P+3工餐2P4（X 表示巳，ck, Br2）P4+20HNC）3（濃）-IH3PQ4+2ONO2 t +4H2OC+2F=CF4C+2C1=CCU占饞占饞C+（）2（足量I*工。22C+OK少量式Uc（）C+CO k=2C（）C+H2（T=CO+H£ 生成水煤氣）2C+SK）當=Si+2C（）（制得粗硅）Si（粗）+2Clk=SiC

6、h（SiCL+2H2=Si（純）+4HCl）Si（粉）+（）金暨Si（）2Si+C直且SiC（金剛砂）Si+2Na()H+H2C)=Na2Si()5+2H2 T(Si+2OH- +H2O= SiO； +2H21)3.岐化反應C12+H2（）=HC1+HC1C）（加堿或光照促進歧化：（C12+H2O=H+C1-+HC1O）Cl>+2Na()H=NaCI+NaCK)+HX) «a2+2Ca(OH)2=CaCl2+Ca(ClO)2+2H2O3cl2+6KOH(濃會 5KCI+KCIQ3+3H2Q(O2+2OH - =C1 - +C1O - +H2O) (CI2+2OH - =ci -

7、 +C1O -+h2o)(3C12+6OH-=5C1 - +CIO3 - +3H2O)3S+6Na()H=2Na2S+Na2S()A+3H2()(3S+6OH-+SO32 -+3H2O)(4P+3OH-+3H2O=PH31 +3H2P(V)4P+3KOH(濃)+3H2()=PH3 +3KH2P()211P+1 5CuSO4+24H2O=5Cu5P+6H3PO4+1 5H2SO43C+CaO=SCaC2+C()1 3C+SK)JfciC+2co t二、金屬單質(Na,Mg,Al,FjCu)的還原性2Na+H=2NaH4Na+O,=2NaQ2NaX)+() A=2NaX), 占供z .«

8、 g、2Na+O盧發(fā)4a2。22Na+S=Na2s (爆炸)2Na+2H2O=2Na()H+H212Na+2H2O=2Na+2OH-+H2 t2Na+2NH3=2NaNH2+H> f 2Na+2NH3=2Na*+2NH2-+H2 t熔融占供A4Na+TiCl= 4NaCl+Ti Mg+Cl=MgCl2 Mg+Br=MgBr>2Mg+。其竺 2VgOMg+S-M gS2Cu+S=Cu2S (Cuf只能由單質制備)Mg+2H2(>=Mg()H)2+H212Mg+TiCL=S=Ti+2MgCl2占燃2Mg+CO 2-2Mg()+CMg+H2S=MgS+H2Mg+H2S()4=MgS

9、()4+H21占供2A1+3C1J' 2AlChMg+2RbCbS=MgCl2+2Rb2Mg+Si()=2Mg()+Si(Mg+2H+=M*+H2 t)4A1+3O2=2A12O3 （常溫生成致密氧化膜而鈍化，在氧氣中燃燒）4Al（Hg）+3O2+2xH2O=2（Al2O3.xH2O）+4Hg （鋁汞齊）高溫AbC)5+3Mn2A1+Cr2（）i-Al2O；+2Cr （鋁熱反應）2A1+Fc2C)a=A12()3+2Fc高溫2Al+3Fc()A12()5+3Fc2Al+6HCl=2AlCh+3H212A1+3H>SC)4=A12(SC)4)3+3H2 2A1+61T=2A13+3

10、H2 f2A1+6H4=2A13+3H212A1+6H2s()式濃)=A12(S(L)3+3S()2+6H2。A1+4HN(X稀)=A1(NO3)3+NQ f +2匕。 2Al+2Na()H+2H2O=2NaAlO2+3H2 (AIR在冷，濃的HfOHNQ；中鈍化) A1+4H+NO3 - =A13+NO t +2H2O 2A1+2OH - +2H2O=2A1O2" +3H212Fc+3Br2=2FcBr5占燃3Fc+2()： =Fc/)42Fc+（）：2Fu。（煉鋼過程）Fc+I=FcbFc+S=FcS (FcS既能由單質制備,又能由高干制備)3Fu+4H2O(©q=Fq

11、O4+4H2 fFc+2HCI=FcC12+H> Fe+2H+=Fe2+H21Fc+CuC12=FcC12+CuFe+C0=F*+Cu JFe+SnC14=FcC12+SnCi2(鐵在酸性環(huán)境下，不能把四氯化錫完全還原為單質相Fu+SnCh=FuCb+Sn JFe+Sn2+=Fe2+Sn J三.非金屬氧化物(HF,HC1,H2O,H2S,NH3)金屬氫化物(NaH)1、還原性：4HCl(?)+MnO=MnC12+Cl> t +2H2O4H+2Q - +MnOrMn2+a21 +2H2O4HC1(濃)+Pb。金事bCb+Cb T +2H2。4H*+2C1 - +PbO2=Pb2+Cl

12、2 f +2H2O4HCKB+。蘆=2CL+2Hq16HCl+2KMnO4=2KCl+2Mna2+5Cl2 t +8H2O16 H+10a+2MnO4- =2Mn2+5Q21 +8H2O6HCI+KCK)3=KC1+3C121 +3H>()6H+5C+C1O3-=3C12 t +3H2O14HCl+K2Cr2O7=2KCI+2CrCh+3C121 +7H2()14Hf+60 - +Cr2O72" =2Cr3"+5Cl21 +7H2O2H2O+2F2=4HF+O22HC1+F2=2HF+C"F2 氣與 HQ、HBr、HR H>S. NH、氣體不能共存)2

13、HBr+C12=2HQ+Bm (CL 氣與 HBr、HL H2S. NH 氣體不能共存)占饞占供 ,2H2S+3OK足量)八 2S()2+2Hq2H2s+()式少量:二旦=2S 1 +2H.O2H2S+S()2=3S I +2H.OHf+HfQ4濃)=S I +SO2 f +2H2O3H2s+2HNOK稀)=3S 1 +2NO +4H2()3H2S+2H+2NO3-=3S | +2NO t +4H2O5H2S+2KMn()4+3H2SC)4=2MnS()4+K2S(+5S I +8H>C)5H2S+2MnO4- +6H+=2Mn2+5S J +8H2O3H£+匕(>2。?

14、+4H2so4=Cr2(S()03+K2so<+3S I +7H2C)3H2S+Cr2O72 - +8H*=2C?+3S J +7H2OH2s+4Na2()2+2H2O=Na2so4+6NaOHH2S+4Na2O2+2H2O=8Na+ SO f+ OH-加執(zhí)2NH3CuO=3Cu+N2+3H2()2NH5+3C12=N2+6HC18NHs+3C12=N2+6NH4C1NHs+NaN()2+HCl=NaCl+N2 +2H.ONH3+NO2-+H+=N21 +2H2O占鑼催化劑4NH3+3(M 純氧蘭經=2、2+6比。4NH3+5()2=4NO+6Hg4NHa+6NC)=5N2+6H2O (

15、用氨清除 NO)NaH+H2O=NaC)H+H2 (生氫劑)NaH+H2O=Na+OH-+H2 f 加初 4NaH+TiCLF=Ti+4NaCl+2H> f CaH2+2H2()=Ca()H)2 +2H2 f2、酸性：4HF+Si()2=SiF4+2Hq (可測定礦樣或鋼樣中Si(%的含星、玻璃雕刻)4HF+Si=SiF4+2H2 2HF+CaCl2=CaF2+2HCIH2s+Fu=F<rS I +H2 fH2S+CuCb=CuS I +2HQ (弱酸制強酸的典型反應)H2S+Cu2+=CuS I +2H+HS+2AgN()3=Ag2s I +2HNO3H2s+2Ag+=Ag2S

16、! +2H+H2S+HgCl2=HgS I +2HC1H2S+Hg2*=HgS | +2H+H2S+Pb(NC)5)2=PbS 1 +2HN(% (鉛試紙檢驗空氣中 H>S)H2S+Pb2"=PbS I +2H+H2s+2Ag=A建S+H? t (銀器在空氣中變黑的原因)2NH5()+2Na=2NaNH2+H2 (NaNH2+H2O=NaC)H+NH3 )3 .NH3的堿性：NHs+HX=NH4X(X: F、Cl、Brx h S)NH+HN()<=NH4N(3NH3+H+=NH4*2NH5+H2SC)4=(NH4)2S()4nh3+h+=nh4+NHs+NaCl+H2()

17、+C()>=NaHC()5+NH4Cl (侯德榜制堿：用于工業(yè)制備小蘇打，蘇打)NHs+H2S=NH4HS4.不穩(wěn)定性：NH3+H2S=NH4+HS-?hf=b5=h +fZrirE 十e22HCh®皿Hz+Cb2H2O2=2H2O+O2加熱h2s=h>+s催化劑2NH<<-N2+3H2四.mE金屬氧化物0。3、S©2、N2。、NQ、N2C）r N（）2、N2Q4、N（）Ss C。、C（）2、Si（）2、P2Q3、P2O5、O2O、02。3、。2。5、O2O7、CIO1、低價態(tài)的還原性：（S（）2、CO、NQ）2so2+O2+2HzO=2H與。（這是

18、S（%在大氣中緩慢發(fā)生的環(huán)境化學反應）催化劑2S5+()2SO3SQ2+NO2=SQ+N。S()2+Cb+2H2C)=H2S()4+2HCI S()2+Br2+2H2()=H2S()4+2HBr S()2+l2+2HK)=H2so4+2H 2NO+()2=2N()2NO+NO2+2Na()H=2NaN()2NO+NO2+2OH -=2NO2-C12+SO2+2H2O=4H+SO42 - +2C1Br2+SO2+2H2O=4H+SO42 - +2BfI2+SO2+2H2O=4H+SO42 - +21 -（用于制硝酸工業(yè)中吸收尾氣中的N。和N（h）2CQ+Q成暨2c（力加執(zhí)CO+Cu（）=2u+C

19、（）2加熱催化劑、3C()+Fc2()52Fu+3c。2C()+H>()7CO2+H22、氧化性:S()、+2H£=3S+2H4) 加執(zhí)S()3+2K1=K>S(53+I2N()2+2Kl+H2()=NC)+b+2I<()H (不能用淀粉KI溶液鑒別溟蒸氣和NOj) 4NO2+H2s=4NO+S()3+H2。加熱力口執(zhí)2N()2+Cu=4Cu()+N2N2()+Zn=W=ZnQ+N2加執(zhí)Si()2+2M g=2Mg( )+SiSOi+H2O=2H*+SO?"CO2+2V屋暨：2MgO+C （CO?不能用于撲滅由Mg,Ci,Ba,Na,K等燃燒的火災）Si(

20、>>+2H 3=Si+2Ho3、與水的作用：S(),+H4)=H,S(%SO1+H2()=H2S()43N()2+H2O=2HNO5+N() (NO?不是硝酸的酸酊) N2O5+H2()=2HNO3N2O5+H2O=2H+2NO5-PzOs+HaCK 冷水)=2HP(%PC+3Hq(熱水)=2H；P()4 (P2()5極易吸水，可作氣體干燥劑)P2Q5+3H2SOK 濃)=2HR)4+3S()3CO2+H2()=H2co3Cl2O+H2O=2HCK)。2(然+乩。=214。4Cl2O7+H2O=2H+2aO4"4 .與堿性物質的作用：S()2+2NH5+H>()=(

21、NH4)2S()5SO2+(NH4)2SO5+H2O=2NH4HSO32NH4HSO3+H2so4=(NH4)2S(L+2H2O+2SO2 麻酸工業(yè)尾氣處理)S()2+Ca(OH)2=CaS(). I +HK)(不能用澄清石灰水鑒別S(%和方.可用品紅鑒別)SO3+MgP=MgSO4SO3+Ca(OH)2=CaSO4 I +H2OCO2+NH3+H2O=NH4HCO3傕仲的ICO2+2NHK 過量>+HQ=(NH4)2C()3(NH4)2C()5<NH2)2C()+2H2()傕仲濟IICO2+2NH,"皿（NH2”CQ+H2（）（工業(yè)制取尿素）CC)2+2Na()Hil&

22、#174;=Na2CO5+H2() C()2(過量)+NaOH=NaHC()3 CO2+Ca(OH)2(jifi)=CaCO3+H2O 2CO2(id®+Ca(OH)2=Ca(HCO5)2 CO2+CaC()s+H2O=Ca(HC()5)22OH+CO2=CO32- +H2Ooh+co2=hco3-Ca2+2OH- +CO2=CaCO31 +H2OOH-+CO2=HCO3-CO2+CaCO3+H2O=Ca2+2HCO3-C()2(不足)+2NaAK)2+3H2O=2Al()H)3 +Na2cO3CO2+3H2O+A1O2- =A1(OH)3 J +co32-CO2(足)+NaAlO2

23、+2HX)=Al(OH)" +NaHC(%CO2+2H2O+A1O2 - =A1(OH)31 +HCO3-CO2+C6H5()Na+H2O=C6H5()H | +NaHC()5C5+C6H5O-+H2O=C6HQH ! +HCO3-SiOz+CaoSffloiSiQ(煉鋼造渣)SiO2+2Na（）H=Na2Si（）3+HK）（常溫下強堿緩慢腐飩玻璃）Sn+NmCQNgSC+CQ”制取玻陶Si()2+CaCO,CaSi()5+C()2 (制取玻璃)2N()2+2Na()H=NaN(52+NaN(VH2()2NO2+2OH-=NO3- +NO2"+H2ONC)+N()2+2Na

24、()H=2NaN()2+H2()(制取硝酸工業(yè)尾氣吸收)NO+NO2+2OFT=2NO3- +H2O五.金屬氧化物1 .低價態(tài)的還原性：6FeO+O2=2Fe3O4Fc()+4HN(=Fc(N()5)3+NC)2+2H2()FeO+4H+NO3-=Fe5+NO2 t +2H2O2 .氧化性：Na2()2+2Na=2Na2()(此反應用于制翁 N。)YgO, AL。?幾乎沒有氧化性，很難被還原為Vg, AL一般通過電解制Mg和ALFU2O3+3H 金 Fu+3H2（）（制還原鐵粉）Fc5()4+4H =3Fc+4H2OCu()+H =u+H2O2Fe3O4+l 6HI=6FeI2+8H2O+2I

25、22Fe3O4+16H+4I-=6Fe2+8H2O+2I2Fc2()5+Fc 邈3FcO (煉鋼過程中加入廢鋼作氧化劑)FuQ+cBFu+c。(高溫煉鋼調節(jié)C含量)2F«)+S畫型2Fu+Si()2 (高溫煉鋼調節(jié)Si含曷)3 .與水的作用：Na2()+H2C)=2Na()HNa2O+H2O=2Na+2OH -2Na2O2+2H2C)=4Na()H+O2 12Na2O2+2H2O=4Na+4OH - +O21(此反應分兩步：Na2O2+2H2()=2NaOH+H2()2；2H2()2=2H2()+()2 H>O2 的制符可利用類似的反應:BaCh+Hf。4稀)=BaSO4+H2

26、()2)MgO+HK)=Mg()H)2(緩慢反應)4,與酸性物質的作用：Na2()+S()5=Na2SC)4Na2()+C()2=Na2CC)5Mg()+S()1=MgS(Na2()+2HCl=2NaCl+H2ONa2O+2H4=2Na+H2O2Na2O2+2C()2=2Na2co3+6 fNa2O2+H2so4(冷，稀)=1£()4+凡。2Mg()+H2SC)4=MgS()4+H2OMgO+2H+=Mg2+H2OA12O5+3H2SO4=A12(SO4)3+3H2OA12O3+6H+=2A1+3H2OAl2O5+2NaC)H=2NaAl()2+H2C)(AI2O3 兩性氧化物)A1

27、2O3+2OH-=2A1O2"+H2OFc()+2HC1=FcC12+H2OFeO+2H*=Fe2+H2OFc2O5+6HCl=2FeCl5+3H2OFe2O3+6H+=2Fe3+3H2OFC5O4+8HCl=FcC12+2FeCls+4H2OFe3O4+8H+=2Fe3+Fe2+4H2O六.含氧酸1、氧化性：4HC1O5+3H2S=3H2SO4+4HC1CIO3 - +3H2S=6H+SO42 - +C1 -HCI(VHI=HI()3+HC1cio3-+i-=io/+a-3HCK)+HI=HIO5+3HCI3HC!O+r=IC>3 - +3H+C1 -HC1O+H2so3=H

28、2S(L+HC1HC1O+H2SO3=3H+SO42 Cl -hck)+h2o2=hci+h2o+o2 tHClO+H2O2=H+a - +H2O+O21(氧化性:HCO>HC1O2>HC1Q3>HC1Q$,但濃，熱的HC1O,氧化性很強)2H2S(X 濃)+C=C()2 1 +25。2 1 +2Hq2H2soM 濃)+7=3S()21 +2H.OH£()4+Fu(A1)室溫下鈍化6Hf。式濃)+2F金FU2(S()4)3+3S()2 t +6H2()2H2so式濃)+Cu=uSO4+SO2 +2H2OH£OK濃)+2HBr=S()2 1 +Br>+

29、2H2()(不能用濃硫酸與 NaBr 制取 HBr) H2S(W)+2HI=SO2 t +L+2HX)(不能用濃硫酸與 Nai 制取 HI)H2so4(稀)+Fu=FuS(L+H2 t2H+Fe=Fe2+H2 tH2so3+2H2s=3S I +3HQ4HNOX 濃)+T=C。2 f +4N()2 1 +2HQ6HN()3(濃)+S=H2so4+6NO2 +2H2()5HN()3(濃)+P=H()4+5NO2 t +曰()5HN(%(稀)+3P+2H2(43HT(L+5N() f5H，+5NO；+3P+2H2O=3H3Po4+5NO t6HN(%濃足)+Fu=Fu(N()3)3+3NO2 t

30、+3H.O4HN()；(濃)+Fu(足)=Fc(N(“2+N()2 1 +2H?O(先得 Fc”,在 Fc 過量時再生成 F+的鹽)4HN(X(稀足)+Fu=Fu(NO3)3+NO t +2乩04H4+NO/+Fe=Fe3*+NO t +2H2O 30HN(55+8Fc=8Fc(NO5)3+3N2O +15H.O30 H+6NO3-+8Fe=8Fe3+3N2O t +15H2O36HN()5+1OFc=1OFc(N()3)5+3N2 +18H2()36H+6NO3- +10Fe=8Fe3+3N2 t +18H2O30HNO5+8Fe=8Fc(NO3)5+3NH4NO5+9H2O30 H+3NO

31、3- +8Fe=8Fes+3NH4+9H2O4Zn+10HN(%(稀)=4Zn(N()3)2+NQ +5H2O4Zn+10H*+2NO5 =4Zn24，+N2O t +5H2O4Zn+10HN(%(稀)=4Zn(N()3)2+NH4N()3+3H2()4Zn+10H*+NO3- =4Zn2*+NH4+5H2O2、還原性:H2SO5+X2+H2O=H2SO4+2HX (X 表示 Cl2,Br2J2)H2SO3+X2+H2O=4H+SO?+X -2H2so3+O2=2H2so42H2so3+C>2=4H+SO/H2S(VH2O2=H2S()4+H2C)H2SO3+H2O2=2H*+SO42-

32、 +h2o5H2so3+2KMnQ4=2MnSC)4+K2so4+2H2SC4+3H2O5H2SO3+2MnO4- =2Mn2+4H+3SO42 - +3H2O H2SC)5+2FcC13+H2O=H2SO4+2FcC12+2HCIH2SO3+2Fe3+H2O=4H+2Fe2+ +SO42-3.酸性：H2S()4()+CaF2 金=CaS(L+2HF ：（不揮發(fā)性酸制取揮發(fā)性酸）H2S()4()+NaC=NaHS()4+HCl （不揮發(fā)性酸制取揮發(fā)性酸）Hf (%(濃)+2NaCI qNa2sQ4+2HC"（不揮發(fā)性酸制取揮發(fā)性酸）HS()4(濃)+NaNo4NaHS()4+HN()

33、3 t（不揮發(fā)性酸制取揮發(fā)性酸）3H2soM 濃)+6。()4)金3CaS(L+2HTO4（強酸制弱酸酸）2H2S(L(濃)+00()4)會2CaSO4+Ca(HPO02（工業(yè)制磷肥）3HN()3+AgI()4=HT(L+3AgNO33H+Ag3PC>4=H3Po4+3Ag*2HNO3+CaCO3=Ca(NOA)2+H2O+CO2 t2H+CaCO3=Ca2+H2O+CO21(用HNOq和濃H2s(%不能制售H2s,H, HBr,S（L等還原性氣體）4HsPO4+Ca3(PO4)2 =3Ca(H2PO4)2（重鈣）H3Po式濃)+NaBr = NaH>PO4+HBr f（不揮發(fā)性酸

34、制取揮發(fā)性酸，磷酸是非氧化性酸）H3P（）4 （濃）+NaI=NaH2PO4+HI 4 .不穩(wěn)定性：2HCK）理皇ZHCl+Oz t （保存在棕色瓶中）4HNON。2 f +。2 t +2總。（保存在棕色瓶中）h2sc）Ah>c）+so2 t （在加熱或酸性條件下分解）h2co=h2（）+cc）2 1 （在加熱或酸性條件下分解）H4SiO=H2Si()5+H2()H2Si()5 = Si()2 I +H2()HBQaEO+S I +SO2 t (在加熱或酸性條件下分解)七.堿1 .低價態(tài)的還原性：4Fe(OH)2+O2+2H2O=4Fe(OH)32 .與酸性物質的作用：2NaOH+S(h

35、(少量尸=Na2sos+HK)OH - +SO2=SO32 - +H2ONa( )H+S(L(足)=NaHS( %OH+SO2 況)=hsc>3-2Na()H+Si()2=Na2Si(VH2OOH-+SiO2=SiO32 -+H2O2Na()H+AU)3=2NaAl()2+H2O2OH+A12O5=2A1O2 -+H2O2KQH+CL=KC1+KC1Q+H 4) CI2+2OH =C1 - +C1O - +H2ONa()H+HCl=NaCI+H>OH+OH' =H2ONa()H+H2s(足)=NaHS+H2。OH-+H2S=HS-+H2O2NaQH+H2s(少量=Na2s+

36、2匕。2OH - +H2S=S2 - +2H2O3NaOH+AlCh=Al(OH)" +3NaCl3OH-+AP+=a1(oh)3 iNa()H+Al(C)H)3=NaAK)2+2H2O(AlCh和 A1(QH)3 哪個酸性強？)OH -+A1(OH)3=A1O2 - +2H2OCa(OH)+2NH4cd26Q2+2NH31 +2HK)(實驗室制 NH5)Na()H+NH4Cl=NaCl+NH3 +H2OMg()H)2+2NH4Cl=MgCl2+2NH3 - H2C)(A1(C)H)3+NH4C1 不溶解)Ba(OH)2+H2SO4=BaS()4 I +2HK)2H+2OH * +B

37、a2+SO42 * =BaSO4 I 2H2。3 .不穩(wěn)定性：Mg(OH)=Mg()+H2C)2A1()H)<=A12O3+3H2O2Fu(OH)金 FU2O3+3H2OCu()H)=Cu()+H2O2AgOH=Ag2()+H2O八、鹽2Fe3+Fe=3Fe2*1 .氧化性：（在水溶液中）2FcC15+Fc=3FcC122FcCl5+Zn(/k<：i=2FcCl>+ZnCl2FcCh+Ag=FcC12+AgCl I Fu2(S(L)3+2Ag=FuS(L+A3SO4 2FcC15+H2S=2FcC12+2HC1+S 2FcCl5+2I<I=2FcC12+2KCl+l2F

38、uC12+Mg=Fu+MgCL2FcCh+Cu=2FuC12+CuC12(用于雕刻銅線路版)2Fe3+Cu=2Fe2+Cu2+2Fe3+Zn=2Fe2*+Zn2*2Fe+a+2Ag=2Fe2+2AgCl I(較難反應)Fu(N()3)3+Ag不反應2R3+H2s=2Fe2+2H+S J2Fe3+2F=2Fe2+I2Fe2*+Mg=Fe+Mg2+NaN()2+NH4Cl=NaCl+N2 t +2比。(實驗室制氮氣)NH4+NO2=N2 f +2H2O2 .還原性：2FuCL+3Q2=2FuCh (在水溶液中不需力口熱)2Fe2+3Cl2=2FeJ+6a3N町S+8HNOK稀)=6NaN()3+2

39、N() f +3S+4H.O3S2-+8H+2NO；=2NO t +3S+4H2O3Na2so3+2HNOK稀)=3Na2S(%+2NO f +HQ35。3>+2H+2NO1=3SO/+2NO t +H2O2Na2s(%+O2=2Na2s(L (Na2S()5 在空氣中易變質)Na)S(%+但NaS(入Na2S+Cb=2NaCl+S I (在水溶液中)S"+C12=2Cr+S |3 .與堿性物質的作用：Ca()H)2+CuS()4=Cu(OH)2 I +CaSC)4 I (波爾多液)MgCU+2NHs H2()=Mg()H)21 +2NH4C1M*+2NH3 H2O=Mg(OH

40、)2 1 +2NH/AICI3+3NH3 h2c)=ai(oh)3 ! +3NH4C1AP+3NH3 H2O=A1(OH)2 I +3NH/FuCh+3NH? - H2O=Fc()H)s I +3NH4CIFe5*+3NHS - H2O=Fe(OH)3 J +3NH/CuS(L+2NH3 氏。丁足)=Cu(QH)2 I +(NH4)2S()4Cu2*+2NH3 H2O=Cu(OH)2 I +2NH/Cu(OH)2+4NH； H2O=Cu(NH3)4()H)2+4H2OCu(OH)2+4NH3 H2O= Cu(NHa)42+2OH +4H2O 銅氨溶液CuS()4+4NH5 匕。(足)=Cu(N

41、HJ4so4+4匕0總方程式Cu2+4NH3 - H2O=Cu(NH3)42+4H2O銅氨溶液AgNOs+NH, - H2O=Ag()H 1 +NH4NOS2AgOH=Ag2O(灰黑色)+H2OAg2O+4NH3 - H2O=2Ag(NH3)2+2OH +3H2O 銀氨溶液AgNOs+2NH3 - H2O=Ag(NH3)2N()3+2H2()Ag+2NH3 - H2O=Ag(NHj2+2H2?？偡匠淌絑nS(L+2NH.3 -凡。(不足)=Zn(OH)21 +(NH4)2S()4Zn2>+2NH3 - H2O=Zn(OH)21 +2NH/Zn(OH)2+4NH3 H2()=Zn(NH1)

42、4()H)2+4H2C)ZnS()4+4NH5 -凡。(足)=Zn(NH04so4+4H2OZn2*+4NH5 - H2O=Zn(NHj4產+4均0總方程式PO/+2H+=H2PO4 hpo?+h+=h2po4' H2PO4+H'=H3PO4C(V+H+=HCOjHCO3 +H*=CO2 t +H2O4 .與酸性物質的作用：強酸制弱酸，或不揮發(fā)性酸制揮發(fā)性酸Na<P()4+2HCl=Na2HP()4+2NaClNa2HP()4+HCl=NaH2P()4+NaClNaH2PO4+Ha=H3PO4+NaaNa2CO5+HCl=NaHC()3+NaClNaHaVHCl=NaCl

43、+H2O+CO2 t3Na2C(V2AlCl5+3H>O=2Al()H)31 +3C()2 t +6NaCl (物質之間的雙水解反應) 3CO32-+2A13+3H2O=2A1(OH)3 j +3CO2 t3Na2co3+2FuCh+3Hq=2Fu(OH)31 +3CO2+6NaCl (物質之間的雙水解反應) 3CO32'+2Fe3*+3H2O=2Fe(OH)3 J +3CO21（物質之間的雙水解反應）（物質之間的雙水解反應）（物質之間的雙水解反應）（物質之間的雙水解反應）3NaHC()5+AlCls=Al(C)H). J +3C6 13HCOf+Al=2Al(OH)3 ； +3

44、CO2 f 3NaHCC)3+FcCl5=Fe(OH)51 +3C(h t3HCO；+Fe=2Fe(OH)3 J +3CO21 3Na2S+A12(S()4)3+6H2O=2Al(C)H)3 I +3H2s 3S2-+2A15*+3H2O=2A1(OH)31 +3H2s 1 3NaAl()2+AlCh+6Hq=4A1()H)3 J +3NaCl3A1O/+A13*+6H2O=4A1(OH)3 J3NaAK)2+FcCh+6H2O=3Al()H)vl +Fc(C)H)3 J +3NaCl3A1O2-+Fe+6H2O=3A1(OH)3 j +Fe(OH)3 JNaAlO2+NH4Cl+2H2C)=

45、Al(C)H)3 1 +NH3 HX)+NaClA1O2-+NH4+2H2O=A1(OH)31 +NH3 H2ONa)C(入+HQ+CO產=2NaHCO、CO32-+H2O+CO2=2HCO；Na2CO5+H2O+2SO2=2NaHS()5+a)2 1 (1：2)CO32-+H2O+2SO2=2HSO；+CO2 t2Na2cOK足)+H2O+SO2=Na2so3+2NaHCO3 (CO2 中的 SCh不能用 Na2c。3 洗氣)2COs2-+H2O+SO2=SOs2-+2HCO； (2:1)N&C()3+SO2=Na2s(%+CO2(1:1)CO32-+SO2=SO32-+CO2NaH

46、C(VS()2=NaHS(Va)2 («為 中的 SCh可能用 NaH"% 洗氣)2HCO3-+SO2=2HSO3-+CO22NaHCO5+SO2=Na2SO3+2CO2+H2O2HCO3-+SO2=SO3+2CO2+H2ONa2SiC)<+2HCl=H2SiO5 J +NaCl 或 Na2Si()5+2HCl+H2C)=H4SiO4 j +2NaClSiO3a+2H+=H5O31或 SiO32-+2H*+H2O=H4SiO4 INa2SiO3+CO2+2HQ=H2SiO3 1 +Na2cO3SiO32-+CO2+2H2O=H4SiO4 I +CO5 .鹽與鹽道分解反

47、應Na2SO4+BaCi2=BaSO41 +2NaCl(沉淀不溶于鹽酸、硝酸)SO32r+Ba2*=BaSO4 INa2S()3+BaC12=BaS(). I +2NaQ (沉淀溶于鹽酸，在硝酸中生成新的沉淀,沉淀不消失)SO+Ba2*=BaSO31Na2C（）s+BaC12=BaC（）51 +2Na。（沉淀溶于鹽酸、沉淀消失）CO32-+Ba2+=BaCO31Na2co3+CaCL=CaCO3 1 +2NaCl (NaHCO,不反應)CO32-+Ca2*=CaCO31Ag+Cl =AgCl IAg'+Bd=AgBr 1Ag+F=AgI IAgN()s+NaCl=AgCl I +NaN

48、(55AgN()s+NaBr=AgBr +NaN()5 AgN(+Kl=AgCl | +KN(%Cu2*+S2=CuS !3AgN(VNasP()4=A&P()4 J +3NaN()s 3Ag+PO4>=A&PO4 ICuS(L+Na2s=CuS +Na2S(FcCI.i+3KSCN=Fc(SCN)s+3KC1F，+3SCN-=Fe(SCN)3 (血紅色，用于Fc?+的特性檢驗)6 .不穩(wěn)定性：Na2S2()5+H2S()4=Na2S()4+S | +SO2 T +H2OS2O32'+2H+=S I +SO2 t +H2ONH4C1=XH. t +HC1 tNH4

49、l=NH5 j +HI 12H1H2+12NH4l=NH； t +H2 1 +12 tNH4HCO含NH? f +HQ+CQ2 t2KN()5=2KN()2+<>2 t2Cu(NOa)=2Cu()+4N()2 f +。2 t光昭2AgN()3=42Ag+2NO2 t +6 f (保存在棕色瓶中)5NH4N()a=4N2 1 +2HN(%+9H2O10NH4N()=8N2 t +4N(h t +O2 t +20比。t (硝酸核爆炸反應)2KMnO4=K2MnO4+MnO2+O2 f2KCl()3=2KCl+3()212NaHC()=Na2C()s+H2C)+C()2 TCa(HC()

50、5)A=CaC(55+H2O+a)2 1CaC()；3Ca()+CO2 t MgCOfiMgO+C()2 t九、電離方程式1 .酸的電商(H2s(%、HN()V HCK HBr、HI、H。«、HF、H2S()s. CH£QOH、H2C()V H2S HNO2、C«HQH、HCN、HC1O)H2so4=2H+SOJ 或：H2so4+2H2O=2H£T+S()/HNO3=H+NO；或：HN()<+H2()=Hp+N()；(以下雷同)hci=h+q-HBr=H+Br"H1=FT+1-HR)二;H+HzPO；H2PHP。： ：H+P()；hfh+

51、f-H2so3=H，+HSO ；HSO ； ；H+SO 丁ch3c o()h=ih+ch 3coo-HCO ；HCO ；CO ；h?s；=h+ hs- hs- ；h+ S”HN()2；H+N。；GHQH=H+aH5()律酚不是酸，顯酸性)HCNH+CN'hck)；h-+qq-2H,O=HQ+()H2 .礴的電離(NaCH、KOH. Ba(OH)2. Mg(OH)2 Al(OH)八 NH3 - H2O) NaOH=N/+OH -K()H=K*+OH -Ba()H)2=Ba2+2OH -Mg(OH)2M+2OVi -A1(OH)；=A+3OH- 酸式電離：A1()H).H+A1O；+H2O

52、NH? H2()=NH：+2OH-Ca(QH)2=Ca2+2QH -(澄清石灰水)Ca()HL；C產+2OH -(石灰懸濁液)3 .鹽的電離(NaCR NafOq、NaHSO、Na2s(%、NaHSC)v MgSO4> CaS()4> Al2(SO4)3x CuS(L、 A1C1V AgN(%、CH£OONa、NH4NO3、FuC%、Na2c(%、NaHC(%、N町S、NaHS、NaH2P(%、 Na2HPO4x Na3Po4、KI、NaBr. NaCKX AgCK CaCOa)NaCl=Na*+CrNaS()4=2Na+SO：-NaHSC)4=H+Na+ SO jNa2

53、so3=2Na+SO廣NaHS()3=Na+HS()3(錯誤書寫：NaHS()3=Na-+H-+SO/)MgSO4=Mgz+SOAL(SO4)3=2AF+3SO：-CuSO4=Cu2+SOJ-AlCh=AF+3Q-AgNO3=Ag+N()3CH3coQNa=CHC()O - +Na+NH4N()5=NH4+N()；FuCh=Fu3+3Q-Na2co3=2Na+CO；NaHC()3=N/+ H8 ；(錯誤書寫：NaHCO3=Na+H+ 8 廣)Na>S=2Na*+S2-NaHS=Na-+HS-(錯誤書寫：NaHS=Na*+H+ S2")NaH>P()4=Na+H2P()；Na2HP()4=2Na+HP()(錯誤書寫：Na2HP()4=2Na+H+P()4")Na3P()4=3Na-+PQ；KI=K+I-NaBr=Na*+Br- Nad()=Na+CK)"AgCI；=Ag+Q- (7隹溶、微溶物質在水中發(fā)生微弱電離)CaCg=CP+CO；-(錯誤書寫：CaCO.=Ca2+C()3")CaSO4；Ca2+SO ; (錯誤書寫:CaSO4=Ca2+SO 7) 3、熔融電高熔融MgCi 筌些=M/-+2Q-Na2()®=2Na+()2-A12Of=2Ap-+3O2-Na。旦吧Na-+C十、水解反應1 .單水解