Question

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

Answer 1

Answer: 3.5 mol of oxygen

Explanation:

The unbalanced equation for this reaction is

$\text{NH}_{3}+\text{O}_{2} \longrightarrow \text{N}_{2}+\text{H}_{2}\text{O}$

Balancing this equation,

$4\text{NH}_{3}+3\text{O}_{2} \longrightarrow 2\text{N}_{2}+6\text{H}_{2}\text{O}$

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.