Calculating Maximum Ammonia Mass from Nitrogen and Hydrogen Reaction

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The discussion centers on calculating the maximum mass of ammonia produced from the reaction of nitrogen and hydrogen. The reaction formula indicates that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia. The user initially questioned the presence of the factor of two in the calculation, which is derived from the stoichiometric ratio of the reaction. After some deliberation, the user clarified that the factor relates to this ratio, confirming their understanding. The final calculated mass of ammonia is 8.33 grams.
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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
 
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dont worry everyone i just worked it out, its to do with the ratio
 
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