Can you please answer these questions of mine:- 224 cm3 of ammonia undergoes catalytic oxidation in presence of pt to give nitric oxide and water vapour. Calculate the volume of oxygen required for the reaction. All volumes measures at room temperature and pressure.
Answers
Answered by
45
4NH3 + 5O2 -----pt ---- 4NO + 6H2O
WHEN THE AMOUNT OF AMMONIA IS 224 cm3 THEN LET AMOUNT OF OXYGEN BE X.
4 VOLUMES OF AMMONIA + 5 VOLUMES OF OXYGEN ---------- 4 VOLUMES OF NITRIC OXIDE + 6 H2O
4 VOLUMES OF AMMONIA = 224
1 VOLUME OF AMMONIA = 224/4 = 56
THEREFORE,
5 VOLUMES OF OXYGEN = 56*5 280cm3
THE VOLUME OF OXYGEN IS 280 cm3.
WHEN THE AMOUNT OF AMMONIA IS 224 cm3 THEN LET AMOUNT OF OXYGEN BE X.
4 VOLUMES OF AMMONIA + 5 VOLUMES OF OXYGEN ---------- 4 VOLUMES OF NITRIC OXIDE + 6 H2O
4 VOLUMES OF AMMONIA = 224
1 VOLUME OF AMMONIA = 224/4 = 56
THEREFORE,
5 VOLUMES OF OXYGEN = 56*5 280cm3
THE VOLUME OF OXYGEN IS 280 cm3.
Answered by
6
Answer ⇒ 280 cm³ of the Oxygen required.
Explanation ⇒
Reaction of the Ammonia with the Oxygen in the Presence of the Platinum is
4NH₃ + 5O₂ ----------→ 4NO + 6H₂O
From this Reaction,
4 vol. of the Ammonia requires 5 vol. of the Oxygen.
∴ 1 vol.------------------------------5/4 vol. of the Oxygen.
∴ 224 cm³--------------------------5/4 × 224 cm³ of the Oxygen.
= 5 × 56 = 280 cm³ of the Oxygen.
Hence, the volume of the Oxygen is required is 280 cm³.
Hope it helps.
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