麻烦老师帮忙解答(1)(2)(3)小问并附上详细解答谢谢
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:化学作业 时间:2024/06/24 00:27:44
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麻烦老师帮忙解答(1)(2)(3)小问并附上详细解答谢谢
![麻烦老师帮忙解答(1)(2)(3)小问并附上详细解答谢谢](/uploads/image/z/5266776-48-6.jpg?t=%E9%BA%BB%E7%83%A6%E8%80%81%E5%B8%88%E5%B8%AE%E5%BF%99%E8%A7%A3%E7%AD%94%EF%BC%881%EF%BC%89%EF%BC%882%EF%BC%89%EF%BC%883%EF%BC%89%E5%B0%8F%E9%97%AE%E5%B9%B6%E9%99%84%E4%B8%8A%E8%AF%A6%E7%BB%86%E8%A7%A3%E7%AD%94%E8%B0%A2%E8%B0%A2)
解题思路: 依据图象分析B的质量分数为100%时放热,结合乙的燃烧热是1molB完全燃烧生成稳定氧化物放出的热量计算出B的物质的量,根据摩尔质量概念计算得到;当全部是A物质放热,结合A的燃烧热计算甲的物质的量,计算摩尔质量;
解题过程:
(1)当X=1.0燃油全部是B,根据B的燃烧热计算B的物质的量,n(B)=4414KJ/ 5518KJ/mol =0.8mol;则M(B)=91.2g/ 0.8mol =114g/mol;
当X=0时,燃油全部是A,依据甲的燃烧热计算甲的物质的量=2066KJ /725KJ/mol =2.84mol,则M(A)=91.2g /2.84mol =32g/mol
(2)32n+114n=73; n= 0.5mol ; 参加反应的O2体积是160L-3.2L=156.8L , n(参O2)=7mol,
;产生CO2气体104L-3.2L=100.8L ,n(CO2)=4.5mol,由于燃料是液体,A是甲醇;
0.5mol甲醇产生CO20.5mol,则B产生CO24moL,所以B分子中又8个C原子,114/14=8…2,
B的分子式是C8H18, N(C)∶N(H)∶N(O)=9:22:1
(3) CH3OH(l)+3/2O2(g)===CO2(g)+2H2O(l) ΔH=-725.8 kJ·mol-1 C8H18(l)+25/2O2(g)=8CO2(g) +9H2O(l)△H= -5518kJ·mol -1
最终答案:(1)32g/mol,114g/mol; (2)A是甲醇; 0B的分子式是C8H18 N(C)∶N(H)∶N(O)=9:22:1 (3) CH3OH(l)+3/2O2(g)===CO2(g)+2H2O(l) ΔH=-725.8 kJ·mol-1 C8H18(l)+25/2O2(g)=8CO2(g) +9H2O(l)△H= -5518kJ·mol -1
解题过程:
(1)当X=1.0燃油全部是B,根据B的燃烧热计算B的物质的量,n(B)=4414KJ/ 5518KJ/mol =0.8mol;则M(B)=91.2g/ 0.8mol =114g/mol;
当X=0时,燃油全部是A,依据甲的燃烧热计算甲的物质的量=2066KJ /725KJ/mol =2.84mol,则M(A)=91.2g /2.84mol =32g/mol
(2)32n+114n=73; n= 0.5mol ; 参加反应的O2体积是160L-3.2L=156.8L , n(参O2)=7mol,
;产生CO2气体104L-3.2L=100.8L ,n(CO2)=4.5mol,由于燃料是液体,A是甲醇;
0.5mol甲醇产生CO20.5mol,则B产生CO24moL,所以B分子中又8个C原子,114/14=8…2,
B的分子式是C8H18, N(C)∶N(H)∶N(O)=9:22:1
(3) CH3OH(l)+3/2O2(g)===CO2(g)+2H2O(l) ΔH=-725.8 kJ·mol-1 C8H18(l)+25/2O2(g)=8CO2(g) +9H2O(l)△H= -5518kJ·mol -1
最终答案:(1)32g/mol,114g/mol; (2)A是甲醇; 0B的分子式是C8H18 N(C)∶N(H)∶N(O)=9:22:1 (3) CH3OH(l)+3/2O2(g)===CO2(g)+2H2O(l) ΔH=-725.8 kJ·mol-1 C8H18(l)+25/2O2(g)=8CO2(g) +9H2O(l)△H= -5518kJ·mol -1