N2 (g) + 3H2 (g) → 2NH3 (g)
1 vol. 3vol. 2vol.
N2: NH3 N2: H2
1:2 1:3
Let the volume of ammonia produced and the volume of consumed hydrogen be x litres and y litres respectively.
200:x 200:y
x=2 ×200 y = 200 ×3
x= 400 y=600
Hence, volume of ammonia produced =400 litres and the volume of excess hydrogen =1000 -600=400 litres
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