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What is the yield of 2,4-dichloronitrobenzene produced from the reaction of 10 grams of nitrobenzene and excess chlorine gas, given that the reaction conditions were optimal and the purity of the final product was determined to be 95%?

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To calculate the yield of 2,4-dichloronitrobenzene, we first need to determine the theoretical yield based on the stoichiometry of the reaction and then account for the 95% purity of the final product.1. Determine the molecular weight of nitrobenzene and 2,4-dichloronitrobenzene.Nitrobenzene  C6H5NO2 : 1 12.01  * 6 + 1 1.01  * 5 + 1 14.01  + 2 16.00  = 123.11 g/mol2,4-Dichloronitrobenzene  C6H3Cl2NO2 : 1 12.01  * 6 + 1 1.01  * 3 + 2 35.45  + 1 14.01  + 2 16.00  = 191.00 g/mol2. Calculate the moles of nitrobenzene used in the reaction.moles of nitrobenzene = mass / molecular weight = 10 g / 123.11 g/mol = 0.0812 mol3. Determine the stoichiometry of the reaction.Since chlorine gas is in excess, the reaction is assumed to go to completion. The stoichiometry of the reaction is 1:1, meaning that 1 mole of nitrobenzene reacts with chlorine gas to produce 1 mole of 2,4-dichloronitrobenzene.4. Calculate the theoretical yield of 2,4-dichloronitrobenzene.moles of 2,4-dichloronitrobenzene = moles of nitrobenzene = 0.0812 molmass of 2,4-dichloronitrobenzene = moles * molecular weight = 0.0812 mol * 191.00 g/mol = 15.49 g5. Account for the 95% purity of the final product.actual yield = theoretical yield * purity = 15.49 g * 0.95 = 14.72 gThe yield of 2,4-dichloronitrobenzene produced from the reaction is 14.72 grams.

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