高中会考化学方程式(总)会考文科班

1、，、非金属单质(F2, Cl2Q2,S,N 2,P,C,Si , H)1、氧化性：3Cl 2+2Fe 2FeCbCl 2+C CuCI?Cl 2+2FeCI2=2FeCI3在水溶液中：Cl 2+2Fe2+=2Fe3+3CI CI2+2Br=2CIBr2CI2+2I =2CI 十l2I Ji, SZ1 "i '#*Cl2+L =2CI- +SJCl 2+2NaBr=2NaCI+Br2Cl 2+2KI=2KCI+12CI 2+Nq S=2NaCI+SjCI2+H2S=2HCI+SJ 水溶液中：Cl2+HS=2Hh+2CI- +SJ Cl 2+SQ+2Hz 0=HSQ+2HCICI

2、2+SO+2HaO=4l1+S- +2CI2、复原性2Q+3Fe 二 FesQN2+3H2催化剂高温高压2NHS+HSQ浓亠3SQ t +2HzQS+6HNQ浓4HzSQ+6NQt +2H2QN2+Q 理 2N0C+C2=2CQC+H=CQ+h生成水煤气3、歧化反响Cl 2+F2Q=HCI+HCICX加碱或光照促进歧化：Cl 2+HQ hT+ci+HCIQCl z+ZNaQHrrNaCI+NaCIQI 2+2QH =CI- +CIQ+"Q制漂白粉：2CI 2+2CaQH=CaCt+CaCIQ2+2H2Q二金属单质(Na,Mg,AI,Fe,Cu )的复原性4Na+(2=2Ns2O2Ns

3、20+42Ns2Q (条件也可以是点燃)2Na+S=NS(爆炸)2Na+2H2O=2NaOH+H2Na+2H2O=2Ns+2OH+" t4AI+ 3C2=2AI 2Q(常温生成致密氧化膜而钝化，在氧气中燃烧)2AI+Fe 2Q 出 Al 2Q+2Fe2AI+ 3FeO±AI 2Q +3Fe2AI+6HCI=2AICI 3+3f t2AI+6H+=2AI 3+3f t2AI+ 3HzSQ=AI2(SO4) 3+3Hz t 2AI+6H+=2AI 3+3H21I I2AI+ 2NaOH2HbO=2NaAIC2+3H21 2AI+2OH +2H2O=2AIO2- +3H213Fe

4、+2Q" FesQ3Fe+4fO(g)亠Fe3Q+4Hz tFe+2HCI=FeC2+H> tFe+2Hh=Fe2+H tO IO IFe+CuC2=FeCI2+CuFe+Cu =Fe +CuJ三非金属氢化物(HF,HCI,H 2O,H2S,NHj)金属氢化物(NaH)1复原性2HBr+Cb=2HCI+Br2(CI 2气与 HBr、HI、HS、NH气体不能共存) 2fS+SO=3S J +2H2O4NH+5Q 4NO+62O2、酸性:4HF+SiQ=SiF4+2liO (玻璃雕刻)3、NH的碱性：NH+HX=N4X(X 代表 F、Cl、Br、I、S)nh+hno>=nhn

5、qnh+h+=nh+2NH+WSQ=(NH) 2SQNH+H+=NH+四非金属氧化物(SQ、SQ、NO NO NO、NG、NQ、NO、COCO、SQ2、RQ、P2Q、CI2Q Cl2Q、CI2Q、CI2Q7、CIO2)1、低价态的复原性：(SO2、CO NO) 催化剂2SQ+Qr 加热二 2SOSQ+CI2+2H2 O=HSQ+2HCICl2+S(O+2H2O=4I1+S(- +2CI-SO+Br2+2H2O=HSOJ+2HBrBr2+S(Q+2F2O=4l1+S(- +2Br-SQ+12+2H2O=HSQ+2HII 2+S(Q+2H2O=4l+S(22 - +2I-2NO+O=2NO3CO+

6、F2Q二 2Fe+3CO2、氧化性:SO+2H2S=3S+2H2O3、与水的作用：SO+“O=HSOSO+fO=HSQ3NQ+H2O=2HN(3+NO(NG不是硝酸的酸酐)4、与碱性物质的作用SQ+Ca(OH2=CaSQj +HO(不能用澄清石灰水鉴别SQ和CQ. 可用品红鉴别)CQ+2NaOlr过量)=NqCQ+H2O 2OF+CQ=cQ - +HOCQ 过量+NaOH=NaHC3£H+CQ=HCOCQ+CaOH过量=CaCO+HzOCa +2oh 十CO=CaCO +H02CQ 过量+CaOH2=CaHCO 2OH+CO=HCO2+-CO+CaCO+HO=CaHCO 2CO+Ca

7、CO+HO=Ca+2HCOSiO2+2NaOH=NSiO3+HO常温下强碱缓慢腐蚀玻璃五金属氧化物2、与水的作用:Ns2O+HO=2NaOHNs2O+HO=2Ns+2OH2Na>O>+2HzO=4NaOH+2OT2Ns202+2H2O=4Ns+4OH+Q f3、与酸性物质的作用:NazO+co=NaCo2NazQ+2CO=2Na2CO+Q fMgO+HBQ=MgSDf OMgO2H =Mg +"OAl 2Q3+3H2SQ=Al 2SOk 3 +3HQAI 20+6=243+3"0Al 2Q+2NaOH=2NaAIQ+HO Al 2Q两性氧化物AI2Q+2OH=2

8、AIO2 +“QFeO+2HCI=FeC|2+HOFeO-2H=Fe2+HOFez Q+6HCI=2FeCI 3 +3F2O Fe2Q+6Hf=2Fd+3HbOFe3Q4+8HCI=FeC2+2FeCI 3 +4F2OFe3Q4+8Hf=2Fe3+Fe2+4HbO六含氧酸1、氧化性:2fSQ浓+C亠CO f +2SQ f +2HbO2H2 SQ浓+A 3SQf +2H2OfSQ+FeAI室温下钝化 6fSQ浓+ 2FFe2SC4 3+3SQ f +602H>SC4浓+C=CuSC+SC f +2H2OH2SC4稀+Fe=FeSC>H> f 2H+Fe=Fe+H f4HN 0

9、浓+CCQf +4NQf +2H2O4、不稳定性：2HCI輕2HCI+C2 f 保存在棕色瓶中4HN0, 4NQf +Qf +2H2C保存在棕色瓶中H2SOH2Q+SQf 在加热或酸性条件下分解H2CQ亠H20+CQf 在加热或酸性条件下分解七.碱 _ 1低价态的复原性：4FeOH2+Q+2H2O=4FeOH32、与酸性物质的作用： 2 -2NaOH+S0少量rNQSO+HCOH+SCSO +H20NaOH+S0足=NaHSQOH+SQ足=HSO2NaOH+Si0=NaSiO3+H2COH+SiQSiQ +"02NaOH+AdC3=2NaAIQ+F2021H+AI 2Q=2AIC2

10、+F203NaOH+AIC3=AIOH3 J +3NaCI 3OH+AI3+=AIOH3 JNaOH+AIOH3=NaAIQ+2H2O0H+AI0H3=AIQ- +2H2OCa0H?+2NHCI - 2CaCb+2NHf +2fO实验室制NH,回忆反响装置是怎么样的NaOH+NCINaCI+NHf +fO3、不稳定性:Mg(OH治 MgO+|O2AI(OH)严 Al 2C3+3H2O2Fe(OH) Fe2Q +3F2O Cu(OH0 CuO+2O八盐1、氧化性：(在水溶液中)2FeCI 3+Fe=3FeCI 22Fe3 +Fe=3Fe22FeCI3+Cu=2FeCI2+CuCb(用于雕刻铜线路

11、版)2Fe3+Cu=2Fe2+CfI I2、复原性：2FeCI2+3CI2=2FeCI3(在水溶液中不需加热) 2Fe2+CI2=2Fe3+2CI-3、与碱性物质的作用：I , AICI 3+3NH HC=AI(OH)j J +3NHCIAI3 +3NH HC=AI(OHb J+3NH+FeCI3+3NH HC=Fe(OH3)J +3NHCIFe3+3NH HC=Fe(OH)J+3NH+4、与酸性物质的作用：强酸制弱酸，或不挥发性酸制挥发性酸(高 沸点制低沸点)Na>C(C +HCI=NaHC3CNaCICO2-+H=HCONaHCC+HCI=NaCI+HO+COfHCO+H=COf +HONa>CO3+H2Q+CC=2NaHCPCGT+HO+C02HCONs2SiQ3+2HCl=JHSiQ3 J +NaCISiQ2-+2H=HSiQ4Ns2SiQ3+C(2+2H2Q=l26iQ3 J +NaCQSiQ32-+CQ+2HQ=HSiQ4 J +CQ-5、盐与盐复分解反响FeCl3+3KSCN=Fe(SCN+3KCIF