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If a sample of benzene (C6H6) is 91% pure and the combustion of benzene by the reaction...

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precious531 | Student, Grade 10 | eNoter

Posted October 14, 2012 at 7:12 AM via web

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If a sample of benzene (C6H6) is 91% pure and the combustion of benzene by the reaction has a 54% yield, how much water is produced from 0.49 mol of a sample of impure benzene?



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anjali989 | Student, Grade 10 | Honors

Posted October 14, 2012 at 8:33 AM (Answer #1)

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The chemical reaction taken place is:

2C6H6+15O2-->12CO2+6H20

2 Moles of benzene:6 moles of water

purity of benzene is 0.49 which is burnt is 91%

so,0.49*91/100=0.04459  moles

yield of reaction=54%

so,amount of water : 0.4459*3=1.3377moles

1.3377*54/100=0.722358

so the water produced is 0.722358

 

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted October 14, 2012 at 7:24 AM (Answer #2)

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The reaction for the combustion of benzene (C6H6) is given by the chemical equation: 2C6H6 + 15O2 --> 12CO2 + 6H2O

2 moles of benzene yield 6 moles of water, or each mole of benzene yields 3 moles of water.

The purity of the 0.49 mole of benzene that is burned is 91%. This gives 0.49*0.91 = 0.4459 moles

The yield of the reaction is 54%. Using this the amount of water produced is 0.4459*3*0.51 = 0.6822

Given that the benzene has a purity of 91% and the yield of the reaction is 54%, 0.49 moles of benzene leads to the production of 0.6822 moles of water.

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