Let's start by writing the equation for this reaction:
#"N"_2 "(g)" + 3"H"_2 "(g)" rightleftharpoons 2"NH"_3 "(g)"#
What we need to do is calculate the moles of hydrogen gas reacting from the given conditions using the ideal-gas equation:
#PV = nRT#
Since we're trying to find the moles, let's rearrange this equation for #n#:
#n = (PV)/(RT)#
In order to use the ideal-gas equation, each component must be in the correct units (#"L", "atm", "mol"#, and #"K"# for volume, pressure, quantity, and temperature, respectively). We need to convert the temperature to Kelvin and the pressure to atmospheres, and we'll use the conversion factor #(1 "atm")/(101.325 "kPa")# and the equation #"K" = ^oC + 273#.
The temperature is
#"K" = 93.0 ^oC + 273 = 366 "K"#
and the pressure
#49.9 cancel("kPa")((1 "atm")/(101.325 cancel("kPa"))) = 0.492 "atm"#
and we're given #12.8 "L"#.
We can now use the ideal-gas equation to solve for the number of moles of #"H"_2#:
#n_("H"_2) = ((0.492 cancel("atm"))(12.8 cancel("L")))/((0.08206 (cancel("L") - cancel("atm"))/("mol"- cancel("K")))(366cancel("K"))) = 0.210 "mol H"_2#
Now we can use the stoichiometric relationships (the coefficients) in the chemical equation to calculate the moles of ammonia that can form (assuming the reaction goes to completion):
#0.210 cancel("mol H"_2)((2 "mol NH"_3)/(3 cancel("mol H"_2))) = 0.140 "mol NH"_3#
We're asked to find the volume of #"NH"_3# that forms from the given conditions, so we'll again use the ideal-gas equation, but this time to solve for #V#:
#V = (nRT)/P#
because we known #n#, #T#, and #P#, but not the volume #V#.
Plugging in known variables, the volume of #"NH"_3# that forms will be
#V = ((0.140 "mol")(0.08206 ("L" - cancel("atm"))/(cancel("mol")- cancel("K")))(366 "K"))/(0.492 "atm") = color(red)(8.55 "L"#
