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Home » 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?
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?
CBSE | Chemistry | Class 11 | Exercise 1 | NCERT | Redox Reactions
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?

Solution:

The reasonable response is as given underneath:

4 \mathrm{NH}_{3(g)}+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}_{(g)}+6 \mathrm{H}_{2} \mathrm{O}_{(g)}

4 N H_{3}=4 \times 17 \mathrm{~g}=68 \mathrm{~g}

5 O_{2}=5 \times 32 \mathrm{~g}=160 \mathrm{~g}

4 N O=4 \times 30 \mathrm{~g}=120 \mathrm{~g}

6 \mathrm{H}_{2} \mathrm{O}=6 \times 18 \mathrm{~g}=108 \mathrm{~g}

Subsequently, N H_{3}(68 \mathrm{~g}) responds with O_{2}(20 \mathrm{~g})

Hence, 10 \mathrm{~g} of N H_{3} responds with \frac{160 \times 10}{68} \mathrm{~g}=23.53 \mathrm{~g} of O_{2}

Yet, just 20 \mathrm{~g} of O_{2} is accessible.

Thus, O_{2} is a restricting reagent.

Presently, 160 \mathrm{~g} of O_{2} gives \frac{120 \times 20}{160} \mathrm{~g} of \mathrm{N}=15 \mathrm{~g} of NO.

Thusly, max of 15 \mathrm{~g} of nitric oxide can be acquired.

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