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?
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.
The chemical reaction taken place is:
2 Moles of benzene:6 moles of water
purity of benzene is 0.49 which is burnt is 91%
yield of reaction=54%
so,amount of water : 0.4459*3=1.3377moles
so the water produced is 0.722358