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?

Question

Infinity Learn · Accepted Answer

$$4NH3(g)+5O2(g)$$⟶$$Pt4NO(g)+6H2O(g)$$  $$4$$×$$17g$$     $$5$$×$$32g$$         $$4$$×$$30g$$         $$68$$ $$gNH3$$ 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 $$120$$×$$1$$×$$20160=15$$ gNO

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?

Unlock the full solution & master the concept.

Ready to Test Your Skills?

free study material