高中化學(xué)方程式大全

1、高中化學(xué)方程式匯編3C12+2PC12+PC13占燃=2FeCl3S+HiN2N22.+盹還原性占燃02S02FeS1低價(jià)態(tài)的還原性：催化劑，叫2SO2+O2+2H2O=2H2SO4無(wú)機(jī)化學(xué)部分非金屬單質(zhì)（F2,Cl2,O2,S,N2,P,C,Si）1. 氧化性:F2+H2=2HF2F2+2H2O=4HF+O2占弊Cl£+Hg-2HC1占燼Pg占燼=PC153C12+2Fe占燃-CuCl;Cl2+2FeCI2=2FeCI32CI2+2NaBr=2NaCI+Br2CI2+2NaI=2NaCI+I2CI2+SO2+2H2O=H2SO4+2HCI（2004北京咼考）占燃20i+3FeJ，.

2、Fe304HSS+Fe乙FeSS+2Cu=£=Cu2S+3H2嚴(yán)壓-皿催化劑dirxail_血訓(xùn)gS+S+6HNO3（濃）=H2SO4+6NO2t+2H2O3S+4HNO3（?。?3SO2+4NOt+2H2O4P+50£山x占撚（X2表示F2,CI2,Br2）占燃2C+少重）2C0C+0"足量）J,co2PX3+X2=PX5C+CO2=2CO（生成水煤氣）aI曰（制得粗硅）啦（粉）+02Si0£Si+2NaOH+H2O=Na2SQ3+2H213. （堿中）歧化CI2+H2O=HCI+HCIO（加酸抑制歧化，加堿或光照促進(jìn)歧化）CI2+2NaOH=NaCI

3、+NaCIO+H2O2CI2+2Ca（OH）2=CaCI2+Ca（CIO）2+2H2O金屬單質(zhì)（Na,Mg,Al,Fe）的還原性4Na+O2=2Na2O占琳2Na£0+J，'2Na£O2占燃2Na+02Ng2Na+S=Na2S（爆炸）2Na+2H2O=2NaOH+H21Mg+H2SO4=MgSO4+H212Mg+022HlgO2A1+3Cli=2A1C134A1+3022A12O32AI+6HCI=2AICI3+3H2t2AI+3H2SO4=AI2（SO4）3+3H2t2AI+6H2SO4（濃、熱）=AI2（SO4）3+3SO2t+6H2O（AI,Fe在冷，濃的H2

4、SO4,HNO3中鈍化）AI+4HNO3（?。?AI（NO3）3+NOt+2H2O2AI+2NaOH+2H2O=2NaAIO2+3H212A1+C“（h二+2Cr言曰4A1+SMnOa，m2AI2O3+3Mn有,回2A1+Fs203'mAlzOs+2Fe3Fe+4H20（g）=Fe04+4H時(shí)Fe+2HCI=FeCI2+H21Fe+CuCI2=FeCI2+Cu非金屬氫化物（HF,HCI,H2O,H2S,NH3）1 還原性：4HC1（濃）+MnO?=MnClj+3時(shí)+2H£016HCI+2KMnO=2KCI+2MnCl2+5CI21+8H2O（實(shí)驗(yàn)室常用2 HzS+SO=3SJ

5、+2HzO2H左+02CJ>量）2S+2H202H2S亠302（足重）=2S0+2H2O2NH+3CI2=N2+6HCI8NH+3CI2=Nz+6NHCI4NHj+3Q（純氧）=2N2+6H2O2NHs+3CuO3Cu+N2+3HR4NH3+5O24N0+6H204NHj+6NO=5N+6HO（用氨清除NO）2. 酸性:4HF+SiQ=SiF4+2HO（HF保存在塑料瓶的原因，此反應(yīng)廣泛應(yīng)用于測(cè)定礦樣或鋼樣中SiO2的含量）HS+CuC2=CuSj+2HCIHS+FeCl2=（不反應(yīng)）3. 堿性：NH+HCI=NHCINH+HNO=NHNO2NH+HSQ=（NH）2SONH+NaCI+H

6、2O+CG>=NaHC9NhlCI（此反應(yīng)用于工業(yè)制備小蘇打，蘇打）4. 不穩(wěn)定性：2H2Q=2HO+Ot非金屬氧化物（這是SO2在大氣中緩慢發(fā)生的環(huán)境化學(xué)反應(yīng)）SO2+Cl2+2H2O=H2SO4+2HCI2NO+O2=2NO22CO+O2=2CO2CO+CuOCu+COjco+h2oSBco£+h23C0+Fe205=2Fe+3C0£2. 氧化性：NO2+2KI+H2O=NO+I2J+2KOH（不能用淀粉.KI溶液鑒別溴蒸氣和NO2）2Mg+C02=2fflg0+C（CO不能用于撲滅由Mg,Ca,Ba,Na,K等燃燒的火災(zāi)）3. 與水的作用：）SO2+H2O=H2

7、SO3SO3+H2O=H2SO43NO2+H2O=2HNO3+NON2O5+H2O=2HNO3P2O5+H2O（冷）=2HPO3P2O5+3H2O（熱）=2H3PO4（P2O5極易吸水，可作氣體干燥劑）4. 與堿性物質(zhì)的作用：SO2+（NH4）2SO3+H2O=2NH4HSO3（這是硫酸廠回收SO2的反應(yīng).先用氨水吸收SO2,再用H2SO4處理:2NH4HSO3+H2SO4=（NH4）2SO4+2H2O+2SO2生成的硫酸銨作化肥,SO2循環(huán)作原料氣）SO2+Ca（OH）2=CaSO3+H2O（不能用澄清石灰水鑒別SO2和CO2.可用品紅鑒別）SO3+Ca（OH）2=CaSO4+H2OCO2+

8、2NaOH（過(guò)量）=Na2CO3+H2OCO2（過(guò)量）+NaOH=NaHCO3CO2+Ca（OH）2（過(guò)量）=CaCO3J+H2O2CO2（過(guò)量）+Ca（OH）2=Ca（HCO3）2CO2+2NaAIO2+3H2O=2AI（OH）3J+Na2CO3CO2+C6H5ONa+H2O=C6H5OHJ+NaHCO3S1O2+COmCaSiOjSiO2+2NaOH=Na2SQ3+H2O（強(qiáng)堿緩慢腐蝕玻璃）聲曰5101 +CaCO3mCaSiO3+CSf5102 +Na2C03mNazSiOa+CO汕金屬氧化物1低價(jià)態(tài)的還原性：FeO+4HNO3=Fe（NO3）3+NO21+2H2O2。氧化性：MgO,A

9、I2O3幾乎沒(méi)有氧化性，很難被還原為Mg,Al.，一般通過(guò)電解制Mg和AI.:（制還原鐵粉）Fe304+4H2血3Fe+4H203與水的作用：Na2O+H2O=2NaOH2Na2O2+2H2O=4NaOH+O214. 與酸性物質(zhì)的作用：Na2O+CO2=Na2CO32Na2O2+2CO2=2Na2CO3+O2Na2O2+H2SO4（冷，?。?Na2SO4+H2O2MgO+SO3=MgSO4MgO+H2SO4=MgSO4+H2OAl2O3+3H2SO4=AI2（SO4）3+3H2OAl2O3+2NaOH=2NaAIO2+H2O（Al2O3是兩性氧化物）FeO+2HCI=FeCI2+3H2OFe2

10、O3+6HCI=2FeCI3+3H2OFe3O4+8HCI=FeCI2+2FeCI3+4H2O含氧酸1. 氧化性：HCIO+H2SO3=H2SO4+HCI溫州市第十五中學(xué)高三化學(xué)備課組PetrosXn.催化劑n匚H2=CHC1催化劑H2AH2CCH3|+3HONO2濃字O2N2Cg鹿+15O2蘭卷12CO2+6H202CH3CH2OH+2Na2CH3CH2ONa+H2fC2H5-OH+H-BrC2H5Br+H2OC2H5OH(液)+3O2(氣)-2CO2(氣)+3H2O(液)催化劑2CH3CH2OH+O22CH3CHO+2H2OA濃h2so4C2H5-O.H+-H-OC2H5-C2H5OC2H

11、5+H2O140C>CH2=CHit+NaBr+HQKaOHj容灌.C2H5-OH+HBr惓化劑CH3CHO+2Ag(WsOHCH3CH2OH確化劑CH3COOHH4+2Ag|+3NH3+H20ACH3CHO+2Cu(OH)2CH3COOH+CU2OJ+2H2O催化劑2CH3CHO+O2-2CH3COOHBrJ+3HBrC17H33C-OCH2oIC17H33COCH0IC17H33C_OCH23H2催化劑加熱，加壓HCIO+H2°2=HCI+H2O+O2(氧化性:HCIO>HCIO2>HCIO3>HCIO4,但濃，熱的HCIO4氧化性很強(qiáng))2H2SO4CS)

12、+c=£=coaf+2SOat+四刃H2SO4+Fe(AI)室溫下鈍化A6H?S04()+2FeFe?(S043+3S0?f+6H2O2H2SO4CS)+CuCUSO4+SOaf+2H£0H2SO4(濃)+2HBr=SO2f+B2f+2H2OH2SO4(稀)+Fe=FeSO4+H24HNO3(濃)+C=CO2f+4NO2f+2H2O6HNO3(濃)+S=H2SO4+6NO2f+2H2O5HNO3(濃)+P=H3PO4+5NO2f+H2O6HNO3(濃、熱)+Fe=Fe(NO3)3+3NO2f+3出04HNO3(稀)+Fe(不足)=Fe(NO3)3+NOf+2H2O8HNO3

13、(稀)+3Fe(過(guò)量)=3Fe(NO3)2+2NOf+4H2O2 還原性：H2SO3+X2+H2O=H2SO4+2HX(X表示Cl2,B2,l2)2H2SO3+O2=2H2SO4H2SO3+H2O2=H2SO4+H2OH2SO3+2FeCl3+H2O=H2SO4+2FeCl2+2HCI3 .酸性:H2SO4(濃)+CaF2=CaSO4+2HFfH2SO4(濃)+NaCI=NaHSO4+HCIfH2SO4(濃)+2NaCI=Na2SO4+2HCIfH2SO4(濃)+NaNO3=NaHSO4+HNO32HNO3+CaCO3=Ca(NO3)2+H2O+CO2f(用HNO3和濃H2SO4不能制H2S,

14、HI,HBr,SO2等還原性氣體)4H3PO4+Ca3(PO4)2=3Ca(H2PO4)2(重鈣)3. 不穩(wěn)定性：2HCIO=2HCI+O2f4HN0a4N02f亠ST十陽(yáng)刃堿1. 低價(jià)態(tài)的還原性：4Fe(OH)2+O2+2H2O=4Fe(OH)32. 與酸性物質(zhì)的作用：2NaOH+SO2(少量)=Na2SO3+H2ONaOH+SO2(足量)=NaHSO32NaOH+SiO2=Na2SiO3+H2O2NaOH+AI2O3=2NaAIO2+H2ONaOH+HCI=NaCI+H2ONaOH+H2S(足量)=NaHS+H2O2NaOH+H2S(少量)=Na2S+2H2O3NaOH+AICI3=AI(

15、OH)3J+3NaCINaOH+AI(OH)3=NaAIO2+2H2ONaOH+NH4CI=NaCI+NH3f+H2OMg(OH)2+2NH4CI=MgCI2+2NH3H2OAI(OH)3+NH4CI(或NH3H2O不溶解)3.不穩(wěn)定性：Mg(OH)2MgO+A2A1(OH),+3H?0A2Fe(0H)3Fe203+3H2O匚U(OH)屯=CuO+HO鹽1 .氧化性：2FeCl3+Fe=3FeCl22FeCl3+Cu=2FeCI2+CuCI2(用于雕刻銅線路版)2FeCl3+H2S=2FeCI2+2HCI+SJ2FeCl3+2KI=2FeCI2+2KCI+I2JFeCl2+Mg=Fe+MgCI

16、22 .還原性：2FeCl2+CI2=2FeCI33Na2S+8HNO3(稀)=6NaNO3+2NOf+3SJ+4H2O3Na2SO3+2HNO3(稀)=3Na2SO4+2NOf+H2O2Na2SO3+O2=2Na2SO43 .與堿性物質(zhì)的作用：MgCI2+2NH3H2O=Mg(OH)2J+NH4CIAICI3+3NH3H2O=AI(OH)3J+3NH4CIFeCl3+3NH3H2O=Fe(OH)3J+3NH4CI4. 與酸性物質(zhì)的作用：Na2CO3+HCI=NaHCO3+NaCINaHCO3+HCI=NaCI+H2O+CO2f3Na2CO3+2AICI3+3H2O=2AI(OH)3J+3CO

17、2f+6NaCI3Na2CO3+2FeCI3+3H2O=2Fe(OH)3J+3CO2f+6NaCI3NaHCO3+AICI3=AI(OH)3J+3CO2f+3NaCI3NaHCO3+FeCl3=Fe(OH)3J+3CO2f+3NaCI3Na?S+Al2(SO4)3+SH20=2Al(OH)3I+3H2St+3Ma2S043NaAIO2+AICI3+6H2O=4AI(OH)3J+3NaCI5. 不穩(wěn)定性：Na2S2O3+H2SO4=Na2SO4+SJ+SO2f+H2ONEjCl=NH3t+HClfANH4HCO?NH3f+HsOt+COif2CuCNO3)32CuO+0汕A2KC1OS=2KC1

18、+NaCOa+H2Of+CO時(shí)Ca(HC03)2=CaCO3+H20+ACaCOg匚aO+CO；fMgCO3=MgO+CO汀有機(jī)化學(xué)部分占燼CH4+202C02+2H20匚田+C12CHCl+HC1CH3CI+Cl£*CHCli+HC1Ahl匚H2CI2+C12*CHC13+HC1CHC13+Cl2CC14+HC1CEi同溫*2H£+C濃硫酸e,丄口aCH3CH2OH*CHpCHgf+HjO17012：占燼CH2=CH2+302>2CO2+2H2OCH2-CH2+fCH2ErCH2ErCH2=CH2+>CH3CH20HnCH2=CH2-催化劑aCaC2+2Hz0C2H2t+Ca(0H)a占堺C2H2+502>4C02+2H2OCH-CH+班2一BrCltCHBrBrCIWCHEr+ErCHBrCHBr?催化叫CH2=CHCi*fCH2-CH(Cl)n濃硝酸一_+HONO2NO2+H2O催化劑A+B2Br2+HBrH2CCCH2CH2CCH3H2N。2+3H2O:CzHj-OH+!CzH5-OH+H-OHOHI+3B2CH3CH?s0H+HOOC