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What is the balanced chemical equation for the preparation of 2,4,6-trinitrotoluene from toluene and nitric acid? Also, calculate the theoretical yield of 2,4,6-trinitrotoluene if 50 grams of toluene and excess nitric acid are used in the reaction.

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The preparation of 2,4,6-trinitrotoluene  TNT  from toluene and nitric acid involves a nitration reaction. The balanced chemical equation for this reaction is:C7H8 + 3 HNO3  C7H5N3O6 + 3 H2OToluene  C7H8  reacts with nitric acid  HNO3  to form 2,4,6-trinitrotoluene  C7H5N3O6  and water  H2O .To calculate the theoretical yield of TNT, we first need to determine the limiting reactant. In this case, we are given 50 grams of toluene and excess nitric acid, so toluene is the limiting reactant.Next, we need to find the molar mass of toluene and TNT:Toluene  C7H8 :  7  12.01  +  8  1.01  = 92.14 g/molTNT  C7H5N3O6 :  7  12.01  +  5  1.01  +  3  14.01  +  6  16.00  = 227.13 g/molNow, we can calculate the moles of toluene used in the reaction:moles of toluene = mass of toluene / molar mass of toluenemoles of toluene = 50 g / 92.14 g/mol = 0.542 molAccording to the balanced chemical equation, 1 mole of toluene reacts to produce 1 mole of TNT. Therefore, the moles of TNT produced will be the same as the moles of toluene used:moles of TNT = 0.542 molFinally, we can calculate the theoretical yield of TNT:mass of TNT = moles of TNT  molar mass of TNTmass of TNT = 0.542 mol  227.13 g/mol = 123.04 gThe theoretical yield of 2,4,6-trinitrotoluene is 123.04 grams.

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