A 2.00 g sample of ammonia (NH3) reacts with 4.00 g of oxygen (O2) according to the equation 4NH3+5O2→4NO+6H2O. How much excess reactant remains after the reaction has stopped ( knowing that NH3 is the excess reactant )?

0.017 moles of ammonia remains after the reaction is stopped.

Explanation:

The reaction is:

4NH₃  +  5O₂  →  4NO  +  6H₂O

The first step is to convert the mass to moles, of each reactant:

2 g .  1mol/ 17g = 0.117 moles of NH₃

4 g . 1mol /32g = 0.125 moles of O₂

Ammonia, states the question, is the excess reactant so we can confirm it,

5 moles of oxygen need 4 moles of ammonia to react (by stoichiometry)

Then, 0.125 moles of oxygen may react to (0.125 . 4) / 5 = 0.1 moles

As we have 0.117 moles of ammonia and we need 0.1 moles.

(0.117 - 0.1) = 0.017 moles remains after the reaction is completed.

If we convert the moles to mass we have:

0.017 mol . 17 g /1mol = 0.289 g

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >

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