题目内容
以CnHmCOOH所表示的羧酸0.1mol加成时需32g溴,0.1mol该羧酸完全燃烧,产生CO2和H2O共3.4mol,该羧酸是( )
| A.C15H27COOH | B.C15H31COOH | C.C17H31COOH | D.C17H33COOH |
0.1mol加成时需32g溴,即n(Br2)=
=0.2mol,说明分子中含有2个C=C键,
A、C15H27COOH中含有2个C=C键,0.1mol该羧酸完全燃烧,产生CO2和H2O共(1.6+1.4)mol=3.0mol,故A错误;
B、C15H31COOH为饱和酸,不含C=C键,故B错误;
C、C17H31COOH中含有2个C=C键,0.1mol该羧酸完全燃烧,产生CO2和H2O共(1.8+1.6)mol=3.4mol,故C正确;
D、C17H33COOH中含有1个C=C键,0.1mol该羧酸完全燃烧,产生CO2和H2O共(1.8+1.7)mol=3.5mol,故D错误.
故选C.
| 32g |
| 160g/mol |
A、C15H27COOH中含有2个C=C键,0.1mol该羧酸完全燃烧,产生CO2和H2O共(1.6+1.4)mol=3.0mol,故A错误;
B、C15H31COOH为饱和酸,不含C=C键,故B错误;
C、C17H31COOH中含有2个C=C键,0.1mol该羧酸完全燃烧,产生CO2和H2O共(1.8+1.6)mol=3.4mol,故C正确;
D、C17H33COOH中含有1个C=C键,0.1mol该羧酸完全燃烧,产生CO2和H2O共(1.8+1.7)mol=3.5mol,故D错误.
故选C.
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