Question #d83e7

Since you can't really have #3.01# molecules of nitrogen gas, I'll assume that you're actually dealing with #3.01 * 10^(23)# molecules of nitrogen gas, the equivalent of

#3.01 * 10^(23)color(red)(cancel(color(black)("molecules N"_2))) * overbrace("1 mole N"_2/(6.02 * 10^(23)color(red)(cancel(color(black)("molecules N"_2)))))^(color(blue)("Avogadro's constant"))#

# = "0.500 moles N"_2#

Now, nitrogen gas and hydrogen gas react to form ammonia according to the following balanced chemical equation

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

Notice that it takes #1# mole of nitrogen gas to produce #2# moles of ammonia, which means that your reaction will produce--keep in mind that the nitrogen gas reacts completely!

#0.500 color(red)(cancel(color(black)("moles N"_2))) * "2 moles NH"_3/(1color(red)(cancel(color(black)("mole N"_2)))) = "1.00 moles NH"_3#

Under Standard Conditions for Pressure and Temperature, #1# mole of any ideal gas occupies #"22.7 L"# #-># this is known as the molar volume of a gas at STP.

You can thus say that your reaction will produce

#1.00 color(red)(cancel(color(black)("moles NH"_3))) * overbrace("22.7 L"/(1color(red)(cancel(color(black)("mole NH"_3)))))^(color(blue)("under STP conditions")) = color(darkgreen)(ul(color(black)("22.7 L")))#

The answer is rounded to three sig figs, the number of significant figures you have for the number of molecules of nitrogen gas.