(1)436kJ/mol+247kJ/mol-2E(H-Cl)=-185kJ/mol,解得E(H-Cl)=434KJ/mol,
故答案为:434KJ/mol;
(2)令氢气和甲烷的物质的量分别为xmol、ymol,根据二者体积与燃烧放出的热量,则:
x+y=
22.4
22.4
571.6
2 x+890.3y=588.05 ,解得x=0.5mol,y=0.5mol,
故氢气的质量为0.5mol×2g/mol=1g,
已知:①CH 4(g)+2O 2(g)═CO 2(g)+2H 2O(l)△H=-890.3 KJ•mol -1
②C(石墨)+O 2(g)═CO 2(g)△H=-393.5KJ•mol -1
③2H 2(g)+O 2(g)═2H 2O(l)△H=-571.6 KJ•mol -1
根据盖斯定律,②+③-①得C(石墨)+2H 2(g)=CH 4(g),故△H=(-393.5KJ/mol)+(-571.6KJ/mol)-(-890.3KJ/mol)=-74.8KJ/mol,
故答案为:1g;-74.8KJ/mol.