C7H6O forms a HSO3 compound of formula C7H7SO4Na in a 1：1 mole ratio.In an experiment it was found that starting from 1.210g of benzaldehyde ,get 2.181g of the hydrogensulphate product.Calculate the percentage yield越详细越好...

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C7H6OformsaHSO3compoundofformulaC7H7SO4Naina1：1moleratio.Inanexperimentitwasfoundthatstartingfrom1.2

C7H6O forms a HSO3 compound of formula C7H7SO4Na in a 1：1 mole ratio.In an experiment it was found that starting from 1.210g of benzaldehyde ,get 2.181g of the hydrogensulphate product.Calculate the percentage yield越详细越好...
C7H6O forms a HSO3 compound of formula C7H7SO4Na in a 1：1 mole ratio.
In an experiment it was found that starting from 1.210g of benzaldehyde ,get 2.181g of the hydrogensulphate product.
Calculate the percentage yield
越详细越好...

C7H6O forms a HSO3 compound of formula C7H7SO4Na in a 1：1 mole ratio.In an experiment it was found that starting from 1.210g of benzaldehyde ,get 2.181g of the hydrogensulphate product.Calculate the percentage yield越详细越好...
91.1%

题目要求的是C7H6O的转化率。
C7H6O + NaHSO3 --> C7H7SO4Na
106..................210
m....................2.181g
解得m = 106 * 2.181/210 = 1.1g
所以有1.1g C7H6O参与了反应
所以转化率 = 1.1/1.210 * 100% = 90.91%

题目告诉你C7H6O反应生成C7H7SO4Na时，原料比为1:1，使用的原料苯甲醛benzaldehyde为1.210克，苯甲醛分子量为106，故为1.21/106=0.0114mol；反应生成的产物为2.181克，产物的分子量为210。按照反应的原料与产物比1:1，应该生成0.0114mol的产物，换算成质量为0.0114*210=2.3972克，题目要你求的是产物的产率，故为2.181/2.3...

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题目告诉你C7H6O反应生成C7H7SO4Na时，原料比为1:1，使用的原料苯甲醛benzaldehyde为1.210克，苯甲醛分子量为106，故为1.21/106=0.0114mol；反应生成的产物为2.181克，产物的分子量为210。按照反应的原料与产物比1:1，应该生成0.0114mol的产物，换算成质量为0.0114*210=2.3972克，题目要你求的是产物的产率，故为2.181/2.3972=90.98%

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