Questions
In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide. that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen?
detailed solution
Correct option is D
4NH3(g)+5O2(g)⟶Pt4NO(g)+6H2O(g) 4×17g 5×32g 4×30g 68 gNH3 reacts with 160 gO21g NH3 reacts with 160×168 g O2∴ 10g NH3 will react with 160×1068=23.5 gO2But available amount of O2 is 20.0 g which is less than the amount which is required to react with 10 g NH3. So, O2 is the limiting reagent and it limits the amount of NO produced.From the above balanced equation,160 g of O2 produces 120 g NO1 g of O2 produces 120×1160 g NO∴20 g of O2 will produce 120×1×20160=15 gNOSimilar Questions
5 moles of x and 9 moles of y will produce
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