[By Lussac’s law]
4NH3 + 5O2 ⟶ 4NO + 6H2O
4 vol : 5 vol ⟶ 4 vol : 6 vol
To calculate the volume of oxygen required.
NH3 : O2
4 : 5
224 : x
⇒ x = 224 × 5\over 4 = 280 cm3
Therefore, volume of oxygen required is 280 cm3.
[By Lussac’s law]
4NH3 + 5O2 ⟶ 4NO + 6H2O
4 vol : 5 vol ⟶ 4 vol : 6 vol
To calculate the volume of oxygen required.
NH3 : O2
4 : 5
224 : x
⇒ x = 224 × 5\over 4 = 280 cm3
Therefore, volume of oxygen required is 280 cm3.
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