0.2mol有机物和0.5molO2在密闭容器中完全燃烧后的产物为CO2、CO和H2O(气),产物经过浓硫酸后,产物质量

2025年04月06日 03:25
有1个网友回答
网友(1):

(1)有机物燃烧生成水10.8g,物质的量为

10.8g
18g/mol
=0.6mol,
令有机物燃烧生成的CO为x,则:
           CuO+CO
 △ 
 
Cu+CO2,气态增重△m
           28g             16g
            x              3.2g
所以x=
28g×3.2g
16g
=5.6g,CO的物质的量为
5.6g
28g/mol
=0.2mol.
根据碳元素守恒可知CO与CuO反应生成的CO2的物质的量为0.2mol,质量为0.2mol×44g/mol=8.8g.
有机物燃烧生成的CO2的质量为17.6g-8.8g=8.8g,物质的量为
8.8g
44g/mol
=0.2mol,
根据碳元素守恒可知,1mol有机物含有碳原子物质的量为2mol,
根据氢元素守恒可知,1mol有机物含有氢原子物质的量为
0.6mol×2
0.2
=6mol
根据氧元素守恒可知,1mol有机物含有氧原子物质的量为
0.6mol+0.2mol+0.2mol×2?0.5mol×2
0.2
=1mol.
所以有机物的分子式为C2H6O.
答:有机物的分子式为C2H6O.
(2)若0.2mol该有机物与足量金属钠作用生成2.24L的H2(标准状况),氢气的物质的量为0.1mol,有机物中含有羟基,该有机物的结构简式为CH3CH2OH.
答:该有机物的结构简式为CH3CH2OH.

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