Calculating Maximum Ammonia Mass from Nitrogen and Hydrogen Reaction

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The discussion focuses on calculating the maximum mass of ammonia (NH3) produced from the reaction of nitrogen (N2) and hydrogen (H2) gases, represented by the equation N2(g) + 3H2(g) → 2NH3(g). The calculation involves determining the number of moles of nitrogen gas, which is then used to find the number of moles of ammonia produced, applying the stoichiometric ratio of 2:1. The final mass of ammonia calculated is 8.33 grams, derived from the molar mass of NH3.

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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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