The key to this problem is the balanced chemical equation that describes this synthesis reaction
#"N"_ (2(g)) + 3"H"_ (2(g)) -> color(red)(2)"NH"_(3(g))#
The balanced chemical equation tells you that when
Now, a mole is simply a very, very large collection of molecules. More specifically, one mole is defined as
#color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"molecules"color(white)(a/a)|))) ->#Avogadro's number
This means that every time you have
So, the reaction will consume
Since the problem doesn't mention the number of moles of nitrogen gas available for the reaction, you can assume that this reactants is in excess, i.e. you have more than you need to ensure that all the moles of hydrogen gas react.
#1.26 color(red)(cancel(color(black)("moles H"_2))) * (color(red)(2)color(white)(a)"moles NH"_3)/(3color(red)(cancel(color(black)("moles H"_2)))) = color(green)(|bar(ul(color(white)(a/a)"0.840 moles NH"_3color(white)(a/a)|)))#
The answer is rounded to two sig figs, the number of sig figs you have for the number of moles of hydrogen gas.