(1)依据反应A中,4mol HCl被氧化,放出115.6kJ的热量,反应的热化学方程式为:4HCl(g)+O2(g)?2Cl2(g)+2H2O(g)△H=-115.6 KJ/mol,
故答案为:4HCl(g)+O2(g)?2Cl2(g)+2H2O(g)△H=-115.6 KJ/mol;
(2)焓变=反应物断键吸收热量-生成物形成化学键放出热量,4HCl(g)+O2(g)?2Cl2(g)+2H2O(g)△H=-115.6 KJ/mol,
4×E(H-Cl)+498-[243×2+4×E(H-O)]=-115.6,得到4×E(H-O)-4×E(H-Cl)=498-486+115.6=127.6
E(H-O)-E(H-Cl)=31.9≈32,H2O中H-O键比HCl中H-Cl键强,
故答案为:32;强;
(3)将所给的三个反应:①+②+③可得总反应:2I-(aq)+O3(g)+2H+(aq)?I2(aq)+O2(g)+H2O(l),△H=△H1+△H2+△H3,
故答案为:2I-+O3+2H+?I2+O2+H2O;△H1+△H2+△H3.