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What is the yield of 2,4-dichloronitrobenzene obtained from the reaction between nitrobenzene and chlorine gas when starting with 10 grams of nitrobenzene and excess chlorine gas, given that the reaction has a theoretical yield of 95%?

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To calculate the yield of 2,4-dichloronitrobenzene, we first need to determine the number of moles of nitrobenzene used in the reaction. We can do this by using the molecular weight of nitrobenzene.The molecular weight of nitrobenzene  C6H5NO2  is:C: 12.01 g/mol  6 = 72.06 g/molH: 1.01 g/mol  5 = 5.05 g/molN: 14.01 g/mol  1 = 14.01 g/molO: 16.00 g/mol  2 = 32.00 g/molTotal molecular weight = 72.06 + 5.05 + 14.01 + 32.00 = 123.12 g/molNow, we can calculate the number of moles of nitrobenzene:moles of nitrobenzene = mass / molecular weightmoles of nitrobenzene = 10 g / 123.12 g/mol = 0.0812 molesThe reaction between nitrobenzene and chlorine gas produces 2,4-dichloronitrobenzene in a 1:1 ratio, so the number of moles of 2,4-dichloronitrobenzene produced is equal to the number of moles of nitrobenzene used.Now, we need to find the mass of 2,4-dichloronitrobenzene produced. To do this, we'll use the molecular weight of 2,4-dichloronitrobenzene  C6H3Cl2NO2 .The molecular weight of 2,4-dichloronitrobenzene is:C: 12.01 g/mol  6 = 72.06 g/molH: 1.01 g/mol  3 = 3.03 g/molCl: 35.45 g/mol  2 = 70.90 g/molN: 14.01 g/mol  1 = 14.01 g/molO: 16.00 g/mol  2 = 32.00 g/molTotal molecular weight = 72.06 + 3.03 + 70.90 + 14.01 + 32.00 = 192.00 g/molNow, we can calculate the mass of 2,4-dichloronitrobenzene produced:mass of 2,4-dichloronitrobenzene = moles  molecular weightmass of 2,4-dichloronitrobenzene = 0.0812 moles  192.00 g/mol = 15.59 gSince the reaction has a theoretical yield of 95%, we need to multiply the mass of 2,4-dichloronitrobenzene by the percentage yield to find the actual yield:actual yield = mass of 2,4-dichloronitrobenzene  percentage yieldactual yield = 15.59 g  0.95 = 14.81 gSo, the yield of 2,4-dichloronitrobenzene obtained from the reaction is 14.81 grams.

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