Answer to Question #11586 in Inorganic Chemistry for swagatika
N2 + 3H2 = 2NH3
n(N2) = m(N2) / M(N2) = 2.8g / 28g/mol =
n(H2) = V(H2) / Vm = 44.8L / 22.711L/mol = 1.97 mol
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
V(NH3) = 2*V(N2) = 2*n(N2)*Vm = 2 * 0.1mol * 22.711L/mol = 4.54 L.
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