Synthesis of Ammonia..

the reaction for producing ammonia is as follows:

${ N }_{ 2 }\quad +\quad 3{ H }_{ 2 }\longrightarrow \quad 2N{ H }_{ 3 }$

$1\quad mole\quad of\quad { N }_{ 2 }+\quad 3\quad moles\quad of\quad { H }_{ 2 }\quad \longrightarrow \quad 2\quad moles\quad of\quad N{ H }_{ 3 }$

$which\quad can\quad also\quad be\quad written\quad as\quad 28g\quad of\quad { N }_{ 2 }+\quad 6g\quad of\quad { H }_{ 2 }\longrightarrow \quad 34g\quad of\quad N{ H }_{ 3 }$

so, we have 18g of hydrogen which is 3 times the minimum required mass of hydrogen(6g)

so, we also require three times the nitrogen to react with 18g of hydrogen that is $(28\times 3)g=84g\quad of\quad { N }_{ 2 }$

so ,out of the 112 g of nitrogen we have, only 84 g reacts. The remaining 28 g(1mole) of nitrogen remains unreacted. Thus, hydrogen in this case is a limiting reagent.

So, mass of ammonia produced = $(84+18)=102\quad g\quad of\quad N{ H }_{ 3 }\quad$

I used 102 because thought that you would have made this mistake.

The correct answer is 101 g.

You have

3.99809 moles of N2

8.9291 moles of H2

The ratio is 1:3. So the limiting reagent is H2 and you will use

8.9291/3 = 2.9764 moles of N2.

2.9764 moles of N2 + 8.9291 moles of H2 -> 5.9528 moles of NH3

5.9528 moles of NH3 = 101 g.

The best thing to do is to accept 101 and 102. Requiring 101 g is picky, only accepting 102 is wrong.

Anyway, thanks for the problem.