Explain what is the maximum number of moles of nh4v3o8

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The compound NH4V3O8 can be prepared according to the following sequence of reactions: N2 + 3H2 -> 2NH3 2NH3 + V2O5 + H2O -> 2NH4VO3 3NH4VO3 + 2HCl -> NH4V3O8 + 2NH4Cl +H2O what is the maximum number of moles of NH4V3O8 that can be prepared starting from 2.5 moles each of N2 and H2. Assume all other reactants are not limiting reagents.

Reference no: EM13298095

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