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?

Asked by Pragya Singh | 1 year ago |  144

##### Solution :-

The balanced reaction is as given below:

$$4NH_{ 3(g) }+5 O_{ 2 \; (g) }\rightarrow 4NO_{ (g) } + 6H_{ 2 }O_{ (g) }$$

$$4 NH_{ 3 }​ = 4 × 17 g = 68 g$$

$$5O_{ 2 }= 5 × 32 g = 160 g$$

$$4 NO = 4 × 30 g = 120 g$$

$$6H_{ 2 }O = 6 × 18 g = 108 g$$

Thus, $$NH_{ 3 }$$​ (68 g) reacts with $$O_{ 2 }$$​ (20g)

Therefore, 10 g of  $$NH_{ 3 }$$ reacts with

$$\dfrac{ 160 \; \times \; 10 }{ 68 }g$$

$$= 23.53 g \;of\; O_{ 2 }$$

But only 20 g of $$O_{ 2 }$$​ is available.

Hence, $$O_{ 2 }$$ is a limiting reagent.

Now, $$160 g \;of \;O_{ 2 }$$​ gives $$\dfrac{ 120 \times 20}{ 160 } g$$ of N = 15 g of NO.

Therefore, max of 15 g of nitric oxide can be obtained.

Answered by Pragya Singh | 1 year ago

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