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What is the optimal ratio of toluene to nitric acid to obtain the highest yield of 2,4,6-trinitrotoluene during the preparation process?

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The optimal ratio of toluene to nitric acid to obtain the highest yield of 2,4,6-trinitrotoluene  TNT  during the preparation process is 1:3. This means that for every mole of toluene, three moles of nitric acid should be used. This ratio ensures that there is enough nitric acid to fully nitrate the toluene molecule at the 2, 4, and 6 positions, resulting in the formation of TNT. However, it is important to note that the reaction is typically carried out in the presence of a strong acid catalyst, such as sulfuric acid, to facilitate the nitration process.
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