(1)将2克液态苯(C6H6)物质的量==mol;完全燃烧生成液态水和CO2,放出83.6KJ的热量,1mol苯燃烧放热=83.6KJ×39mol=3260.4KJ; 反应的热化学方程式为:C6H6(l)+O2(g)→6CO2(g)+6H2O(l)△H=-3260.4KJ/mol, 故答案为:C6H6(l)+O2(g)→6CO2(g)+6H2O(l)△H=-3260.4KJ/mol; (2):①C(s)+O2(g)=CO2(g);△H=-393.5kJ•mol-1 ②2CO(g)+O2(g)=2CO2(g);△H=-566kJ•mol-1 ③TiO2(s)+2Cl2(g)=TiCl4(s)+O2(g);△H=+141kJ•mol-1 依据盖斯定律,①×2-②+③得到:TiO2(s)+2Cl2(g)+2C(s)=TiCl4(s)+2CO(g)△H=-80KJ/mol, 故答案为:TiO2(s)+2Cl2(g)+2C(s)=TiCl4(s)+2CO(g)△H=-80KJ/mol. |