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Ammonia (NH3) gas and oxygen (O2) gas react to form nitrogen (N2) gas and water (H2O) vapor. Suppose you have 2.0 mol of NH3 and 5.0 mol of O2 in a reactor. Suppose as much as possible of the NH3 reacts. How much will be left? Round your answer to the nearest 0.1 mol

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Medunno13

Answer: 3.5 mol of oxygen

Explanation:

The unbalanced equation for this reaction is

[tex]\text{NH}_{3}+\text{O}_{2} \longrightarrow \text{N}_{2}+\text{H}_{2}\text{O}[/tex]

Balancing this equation,

[tex]4\text{NH}_{3}+3\text{O}_{2} \longrightarrow 2\text{N}_{2}+6\text{H}_{2}\text{O}[/tex]

From this, we can tell that for every 4 moles of ammonia consumed, 3 moles of oxygen are consumed.

  • Considering the ammonia, the reaction can occur 2.0/4 = 1/2 a time.
  • Considering the oxygen, the reaction can occur 3/5 = 3/5 a time.

This means that ammonia is the limiting reactant, meaning that 2.0(3/4)=1.5 moles of oxygen are consumed.

So, 5.0-1.5=3.5 mol of oxygen remain.

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