Question #fb8ba

Start by writing the balanced chemical equation that describes your reaction

#"N"_ (2(g)) + color(blue)(3)"H"_ (2(g)) -> color(purpple)(2)"NH"_ (3(g))#

Notice that every mole of nitrogen gas that takes part in the reaction consumes #color(blue)(3)# moles of hydrogen gas and produces #color(purple)(2)# moles of ammonia.

Now, in order to be able to use these mole ratios, you must convert the sample of ammonia from grams to moles. To do that, use the compound's molar mass

#13.91 color(red)(cancel(color(black)("g"))) * "1 mole NH"_3/(17.0305 color(red)(cancel(color(black)("g")))) = "0.81677 moles NH"_3#

Now, in order for the reaction to produce that many moles of ammonia, it must consume

#0.81677 color(red)(cancel(color(black)("moles NH"_3))) * (color(blue)(3)color(white)(.)"moles H"_2)/(color(purple)(2)color(red)(cancel(color(black)("moles NH"_3)))) = "1.2252 moles H"_2#

So, you can say that #1.2252# moles of hydrogen gas reacted to produce your sample of ammonia. To convert this to grams, use the molar mass of hydrogen gas

#1.2252 color(red)(cancel(color(black)("moles H"_2))) * "2.0159 g"/(1color(red)(cancel(color(black)("mole H"_2)))) = color(darkgreen)(ul(color(black)("4.470 g")))#

The answer is rounded to four sig figs, the number of sig figs you have for the mass of ammonia.