Calculating Maximum Ammonia Mass from Nitrogen and Hydrogen Reaction

[alexparker]

The Question is
Ammonia is formed by the reaction of nitrogen and
hydrogen:
N2(g) + 3H2(g) → 2NH3(g)
Calculate the maximum mass of ammonia that
could be produced when 6.00 L of nitrogen gas at
SLC reacts with excess hydrogen gas.

The Answer is
n(N2) = = mol
n(NH3) = 2 × n(N2)
m(NH3) = n × M = = 8.33 g

i don't know why the 2 in the step 2 of the answer is there. I really need to know as i don't understand why it is there, and i cannot ask my teacher as i am on holidays.

Thanks

 

[alexparker]

dont worry everyone i just worked it out, its to do with the ratio

 

[Blueshift5]

for your question. The 2 in step 2 of the answer represents the stoichiometric coefficient of NH3 in the balanced chemical equation. This coefficient indicates the number of moles of NH3 produced for every 1 mole of N2 reacted. In this reaction, 2 moles of NH3 are produced for every 1 mole of N2 reacted. Therefore, in order to calculate the maximum mass of NH3 produced, we must multiply the number of moles of N2 by 2. This is why the 2 is included in the calculation. I hope this helps clarify any confusion. Enjoy your holidays!