Question

30cm of N2 is mixed with 40cm of H2 under suitable condition to produce NH3
1.Write the balanced chemical equation for the reaction
2. Assuming the reaction gives the maximum yield of NH3 calculate
The volume of NH3 produced
The excess of any
reactant gas
The total volume of gaseous mixture after reaction

Answer 1

1. N₂ + 3H₂ → 2NH₃

2. V(N₂) = 30 mL

V(H₂) = 40 mL

n(N₂) $= \frac{30}{22.4 \times 10^{3}} = 1.34 \times 10^{-3} \;mol$

n(H₂) $= \frac{40}{22.4 \times 10^{3}} = 1.78 \times 10^{-3} \;mol$

According to the reaction for each mole of N₂ we need 3 moles of H₂. So, H₂ is a limitting reactant.

n(NH₃) $= \frac{2}{3}n(H_2) = \frac{2}{3} \times 1.78 \times 10^{-3} = 1.19 \times 10^{-3} \;mol$

V(NH₃) $= 1.19 \times 10^{-3} \times 22.4 \times 10^{3} = 26.66 \;mL$

The volume of NH3 produced is 26.66 mL.

$n(N_2)_{used} = \frac{1}{3}n(H_2) \\ = \frac{1}{3} \times 1.78 \times 10^{-3} \\ = 0.593 \times 10^{-3} \;mol$

$V(N_2)_{used} = 0.593 \times 10^{-3} \times 22.4 \times 10^3 = 13.3 \;mL$

ΔV(N₂) = 30 -13.3=16.7 mL (excess of N₂)

The total volume of gaseous mixture after reaction = 16.7 mL + 26.66 mL = 43.36 mL