How many moles of NO are made from mixing 7.2 moles of NH3 and 9.6 moles
of O? Identify the limiting reagent.

216 g of NO

Explanation:

We begin from the reaction:

4NH₃ + 5O₂ →  4NO  + 6H₂O

We determine the limiting reactant with the moles of each reactant:

4 moles of ammonia react to 5 moles of oxygen

Our 7.2 moles of ammonia may react to (7.2 . 5) /4 = 9 moles

It's ok because we have 9.6 moles of oxygen. 0.6 moles still remain.

5 moles of oxygen react to 4 moles of NH₃

Our 9.6 moles of oxygen may react to (9.6 . 4) /5 = 7.68 moles

We only have 7.2 moles of NH₃ and we need 7.68; so there is no enough ammonia and that's our limiting reagent.

Now we determine the moles of product.

4 moles of ammonia can produce 4 moles of NO

Definetely our 7.2 moles, will produce 7.2 moles of oxide.

We convert to mass: 7.2 mol . 30 g/mol = 216 g

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