Answer to Question #11586 in Inorganic Chemistry for swagatika

Question

what will be the volume of ammonia that can be produced out of 2.8g of nitrogen and 44.8l of hydrogen at S.T.P in haber's process

Answer 1

Haber's process:
N2 + 3H2 = 2NH3
n(N2) = m(N2) / M(N2) = 2.8g / 28g/mol =
0.1 mol
n(H2) = V(H2) / Vm = 44.8L / 22.711L/mol = 1.97 mol
The molar
volume of an ideal gas at STP (100 kPa and 273.15K) is 22.711 L/mol
H2 is the
excess reagent, and N2 is the limiting reagent.
Thus, the volume of ammonia
is
V(NH3) = 2*V(N2) = 2*n(N2)*Vm = 2 * 0.1mol * 22.711L/mol = 4.54 L.