Ammonia gas combines with oxygen gas over "Pt " catalyst to produce nitric oxide (NO) and water. If 6.8 g NH₃ gas is taken initially then;

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Number of moles of "NO" produced is 0.4

Total mass of products obtained is 22.8 g

Volume of H₂O produced at 4⁰C (assuming density of water as 1000 kg / m³ is 21.6 $mℓ$

Volume of oxygen gas required at 1 atm and 273 K is 11.2 litre

answer is A, C, D.

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Detailed Solution

$4 {NH}_{3} + 5 O_{2} \to 4 NO + 6 H_{2} O$
$∴$ The volume of oxygen required =11.2 lit
Total mass of products $(NO + H_{2} O)$ is = 22.8 g
The number of moles of NO produced = 0.4

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