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

The compound originally contained KNO₃ solution.

Explanation:

When a mixture which contains either Pb(NO₃)₂ or KNO₃ is added with NaCl solution, no reaction takes place. This confirms the presence of KNO₃ in the solution.

The reason behind the above activity can be attributed to the lower placement of Na (sodium) in reactivity charge than potassium (K). Hence, it cannot displace Potassium (K) from the aqueous KNO₃ solution. Whereas the NaCl reacts with Pb(NO₃)₂ to produce white insoluble precipitate of PbCl₂. This confirms the presence of KNO₃ in solution.