第16章碳族元素习题解答.doc

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1、第16章碳族元素习题解答1．下图为碳元素部分不同存在形态物质之间的转化图，请写出具体的反应方程式。解4CO(g)+Ni(s)®Ni(CO)4(g)CO(g)+Cl2(g)®COCl2(g)CO(g)+S(s)®COS(g)HCOOH(l)CO(g)+H2O(l)CO2(g)+2Ca(s)®C(s)+2CaO(s)2CO(g)+O2(g)®2CO2(g)CO(g)+2H2(g)CH3OH(l)2C(s)+O2(g)®2CO(g)C(s)+O2(g)®CO2(g)Na2C2(s)+2H2O(l)®2NaOH(aq)+C

2、2H2(g)2C2H2(g)+5O2(g)®4CO2(g)+2H2O(l)Al4C3(s)+H2O(l)®3CH4(g)+4Al(OH)3(s)CH4(g)+2O2(g)®CO2(g)+2H2O(l)CH4(g)+4S(l)®CS2(g)+2H2S(g)CS2(g)+3Cl2(g)®CCl4(g)+S2Cl2(l)CS2(g)+S2Cl2(l)®CCl4(g)+6S(s)CH4(g)+NH3(g)®HCN(g)+3H2(g)HCN(aq)+H2O(l)®H3O+(aq)+CN-(aq)Ca(OH)2(aq)+CO2

3、(g)®CaCO3(s)+H2O(l)CaCO3(s)+2HCl(aq)®CaCl2(aq)+H2O(l)+CO2(g)CaCO3(s)+H2O(l)+CO2(g)®Ca(HCO3)2(aq)2．下图为部分硅元素部分不同存在形态物质之间的转化图，请写出具体的反应方程式。解Si(s)+HCl(g)®SiHCl3(g)+H2(g)2CH3Cl(g)+Si(s)®(CH3)2SiCl2(l)SiO2(s)+2C(s)®Si(s)+2CO(g)SiO2(s)+6HF(aq)®SiF62−(aq)+2H+(aq)+2H2O(

4、l)SiO2(s)+2NaOH(l)®Na2SiO3(s)+H2O(g)SiO2(s)+3C(s)®SiC(s)+2CO(g)SiO2(s)+2Na2CO3(l)Na4SiO4(s)+2CO2(g)2SiO44-(aq)+2H+(aq)®Si2O76-(aq)+H2O(l)3．完成下列反应方程式：（1）固体Li2C2与水反应→；（2）SiO2与C反应→；（3）CuO(s)与CO共热→；（4）Ca(OH)2水溶液中通入CO2（两个反应）→；（5）甲烷与熔融的S反应→；（6）SiO2与熔融的Na2CO3反应→；（7）P

5、bO2与浓HCl反应（两个反应）→。解(1)Li2C2(s)+2H2O(l)®2LiOH(aq)+C2H2(g)(2)SiO2(s)+2C(s)®Si(l)+2CO(g)(3)CuO(s)+CO(g)®Cu(s)+CO2(g)(4)Ca(OH)2(aq)+CO2(g)®CaCO3(s)+H2O(l)CaCO3(s)+CO2(g)+H2O(l)®Ca(HCO3)2(aq)(5)CH4(g)+4S(l)®CS2(g)+2H2S(g)(6)SiO2(s)+2Na2CO3(l)®Na4SiO4(s)+2CO2(g)(7)P

6、bO2(s)+4HCl(aq)®PbCl4(aq)+2H2O(l)PbCl4(aq)®PbCl2(s)+Cl2(g)4．通过比较下列化合物间性质的差别，比较C和N两元素的差别：（1）CH4和NH3；（2）C2H4和N2H4。解hydrocarbonshaveverylowboilingpoints,whilethehydrogen-bondinghydridesofcarbonhavemuchhigherboilingpoints.Followingfromthis,theyhavedifferentacid-ba

7、seproperties,bothhydridesofcarbonbeingneutralwhilethoseofcarbonarebasic:NH3(aq)+H2O(l)®NH4+(aq)+OH–(aq)N2H4(aq)+H2O(l)®N2H5+(aq)+OH–(aq)Themostobviousdifferenceisintheircombustions,bothhydrocarbonsproducingcarbondioxidewhilethecarbonhydridesproducecarbongas:CH

8、4(g)+2O2(g)®CO2(g)+2H2O(g)2C2H4(g)+6O2(g)®4CO2(g)+4H2O(g)4NH3(g)+3O2(g)®2N2(g)+6H2O(g)N2H4(l)+O2(g)®N2(g)+2H2O(l)5．对比CO2和CS2，O、S同属16族元素，结构均为直线型。但是CO2的标准摩尔生成焓小于零，而CS2的标准摩尔生成焓却大于零。利用