当5.8kg丁烷完全反应物质的量
=100mol,完全燃烧并生成二氧化碳和液态水时,放出热量为2.9×105kJ,2mol丁烷完全反应放热=5800g 58g/mol
=5800KJ,依据反应物和产物状态标注聚集状态和对应量下的反应热,书写的热化学方程式为:2C4H10(g)+13O2(g)=8CO2(g)+10H2O(l)△H=-5800kJ/mol,已知1mol液态水汽化时需要吸收44kJ热量,①H2O(l)=H2O(g)△H=44KJ/mol;②2C4H10(g)+13O2(g)=8CO2(g)+10H2O(l)△H=-5800kJ/mol;2.9×105KJ 50
依据盖斯定律①×10+②得到2C4H10(g)+13O2(g)=8CO2(g)+10H2O(g)△H=-5360kJ/mol,则得到反应,C4H10(g)+
O2(g)═4CO2(g)+5H2O(g)△H=-2680KJ/mol;13 2
故答案为:2C4H10(g)+13O2(g)=8CO2(g)+10H2O(l)△H=-5800kJ/mol;-2680KJ/mol;