DATA

Mole of N₂ = 6 moles
Mole of H₂ = 8 moles

Asked
a/ i/ Limiting reagent.
ii/ Excess reagent.
b/ Volume of NH₃ Formed.

Solution
Equation for the reaction
N₂(g) + 3H₂(g) ⇋ 2NH₃(g)
1 mole: 3 moles: 2 moles

To find mole ratio for H₂ and N₂
Mole ratio for H₂ = mole reacted / Mole from equation
= 8 moles / 3 moles
= 2.67 mole

Mole ratio for N₂ = mole reacted / Mole from equation
= 6 moles / 1 mole
= 6

a/ i/ The limiting reagent is hydrogen gas (H₂)
ii/ The excess reagent is nitrogen (N₂)

b/ To find the volume of ammonia gas formed

To use mole of limiting with equivalent to ammonia

3 moles of H₂ Ξ 2 moles NH₃
8 moles of H₂ Ξ??

(8 moles × 2 moles) /3 moles
= 5.33 mole of NH₃

Volume = mole × molar volume
= 5.33 mole × 22.4dm³/mole
= 119.47dm³
119.47dm³ of ammonia gas was produced at STP