Let's start by writing the equation for this reaction:
"N"_2 "(g)" + 3"H"_2 "(g)" rightleftharpoons 2"NH"_3 "(g)"N2(g)+3H2(g)⇌2NH3(g)
What we need to do is calculate the moles of hydrogen gas reacting from the given conditions using the ideal-gas equation:
PV = nRTPV=nRT
Since we're trying to find the moles, let's rearrange this equation for nn:
n = (PV)/(RT)n=PVRT
In order to use the ideal-gas equation, each component must be in the correct units ("L", "atm", "mol"L,atm,mol, and "K"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")1atm101.325kPa and the equation "K" = ^oC + 273K=oC+273.
The temperature is
"K" = 93.0 ^oC + 273 = 366 "K"K=93.0oC+273=366K
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"
