Question #cad1e

For starters, the balanced chemical equation

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

tells you that the reaction consumes #color(blue)(3)# moles of hydrogen gas (the #color(blue)(3)# coefficient added in front of hydrogen gas) and produces #color(purple)(2)# moles of ammonia (the #color(purple)(2)# coefficient added in front of ammonia) for every #1# mole of nitrogen gas (we never add a coefficient of #1# in chemical equations) that takes part in the reaction.

Now, the problem tells you that the reaction consumes #0.3# moles nitrogen gas. Since no mention of how much hydrogen gas you have available, you can assume that hydrogen gas is in excess, i.e. you have more than you need to ensure that all the moles of nitrogen gas react.

RIght from the start, you can use the #1:color(purple)(2)# mole ratio that exists between nitrogen gas and ammonia to calculate how many moles of ammonia will be produced

#0.3 color(red)(cancel(color(black)("moles N"_2))) * (color(purple)(2)color(white)(.)"moles NH"_3)/(1color(red)(cancel(color(black)("mole N"_2)))) = "0.6 moles NH"_3#

Now that you know how many moles of ammonia you have, you can use the molar mass of ammonia to convert that to grams.

Ammonia has a molar mass of #"17.031 g mol"^(-1)#, which means that #1# mole of ammonia has a mass of #"17.031 g"#.

You can thus say that the reaction will produce

#0.6 color(red)(cancel(color(black)("moles NH"_3))) * "17.031 g"/(1color(red)(cancel(color(black)("mole NH"_3)))) = color(darkgreen)(ul(color(black)("10 g")))#

The answer is rounded to one significant figure, the number of sig figs you have for the number of moles of nitrogen gas.

So remember, you can go from moles to grams by using the molar mass of the compound.

You always set up the conversion factor as

#"the mass of 1 mole in grams"/"1 mole of the substance"#

When you multiply this conversion factor by the number of moles of the substance, you end up with the mass in grams of your sample.