Final answer:
In the Hoffmann Bromamide Reaction, each mole of benzamide reacts with one mole of KOH and Br2 in a 1:1:1 ratio to form aniline. Therefore, 1 mole of benzamide will consume 1 mole of KOH.
Step-by-step explanation:
To determine how many moles of KOH are consumed in converting 1 mole of benzamide to aniline through the Hoffmann Bromamide Reaction, we need to understand the stoichiometry of the reaction. The Hoffmann Bromamide Reaction transforms an amide into an amine by reacting it with bromine and an aqueous base, typically KOH. The balanced chemical equation for the reaction of benzamide (C6H5CONH2) to aniline (C6H5NH2) is as follows:
Benzamide + KOH + Br2 → Aniline + KBr + CO2 + H2O
This reaction proceeds in a 1:1:1 molar ratio for benzamide, KOH, and Br2 respectively. Therefore, for the conversion of 1 mole of benzamide to aniline, 1 mole of KOH will be consumed. This is assuming that the reaction goes to completion and all reagents are present in stoichiometric amounts.