NH3 (g) + O2 (g) ( ∆ ) ⃗ NO(g) + H2O(l)
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4NH3 + 5O2 --> 4NO + 6H2O
3moles NH3 x (5 O2 / 4 NH3) = 3.75moles O2 required
4moles O2 / (4 NH3 / 5 O2) = 3.2moles NH3 required
NH3 = limiting
3moles NH3 (4 NO / 4 NH3) = 3moles NO
3moles NH3 x (5 O2 / 4 NH3) = 3.75moles O2 required
4moles O2 / (4 NH3 / 5 O2) = 3.2moles NH3 required
NH3 = limiting
3moles NH3 (4 NO / 4 NH3) = 3moles NO