MCQ 11 Mark
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?
- ✓15g of NO
- B20g of NO
- C16g of NO
- D25g of NO
Answer
View full question & answer→Correct option: A.
15g of NO
(a) 15 g of NO
Explanation: The reaction that takes place in ammonia and oxygen is given below:
$
\underset{68 g}{4 NH_3}+\underset{160 g}{5 O_2} \rightarrow \underset{120 g}{4 NO}+6 H_2 O
$
Limiting reagent
160 g of oxygen reacts with 68 g of ammonia
20 g of oxygen reacts with $\frac{68 \times 20}{160}=8.5 g$ of ammonia
Therefore for 20 g of oxygen 8.5 g of ammonia is used. Therefore 1.5 g ammonia is in excess and therefore oxygen is the limiting reagent.
$
160 g \text { of } O_2 \text { produces }=120 g \text { of } NO
$
Therefore, 20 g of $O _2$ produces $=\frac{120 \times 20}{60}=15 g$ of NO.
Explanation: The reaction that takes place in ammonia and oxygen is given below:
$
\underset{68 g}{4 NH_3}+\underset{160 g}{5 O_2} \rightarrow \underset{120 g}{4 NO}+6 H_2 O
$
Limiting reagent
160 g of oxygen reacts with 68 g of ammonia
20 g of oxygen reacts with $\frac{68 \times 20}{160}=8.5 g$ of ammonia
Therefore for 20 g of oxygen 8.5 g of ammonia is used. Therefore 1.5 g ammonia is in excess and therefore oxygen is the limiting reagent.
$
160 g \text { of } O_2 \text { produces }=120 g \text { of } NO
$
Therefore, 20 g of $O _2$ produces $=\frac{120 \times 20}{60}=15 g$ of NO.
