Nitrogen and hydrogen gases react to form ammonia gas via the following reaction:
N2(g)+3H2(g)→2NH3(g)
At a certain temperature and pressure, 1.9 L
of N2
reacts with 5.7 L
of H2
.If all the N2
and H2
are consumed, what volume of NH3
, at the same temperature and pressure, will be produced?
From the balanced equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
First, we need to determine the number of moles of N2 and H2 present:
1.9 L N2 * (1 mol N2 / 22.4 L) = 0.0848 mol N2
5.7 L H2 * (1 mol H2 / 22.4 L) = 0.2545 mol H2
Since N2 and H2 react in a 1:3 ratio, we can see that N2 is limiting reactant as it produces less moles of NH3 which is 0.0563 moles.
Now, we can use this information to calculate the volume of NH3 produced:
0.0563 mol NH3 * (22.4 L / 2 mol NH3) = 0.632 L NH3
Therefore, 0.632 L of NH3 will be produced in this reaction.