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?
The balanced chemical equation for the given reaction is given as:
`4NH_(3(g)) + 5O_(2(g)) -> 4NO_(g) + 6H_2O_(g)`
`4xx 17g 5xx 32g 4xx 30g 6xx18g`
= 68g =160g =120g = 108g
Thus, 68 g of NH3 reacts with 160 g of O2.
Therefore, 10g of NH3 reacts with `(160xx10)/68` g of `O_2` or 23.53 g of `O_2`
But the available amount of O2 is 20 g.
Therefore, O2 is the limiting reagent (we have considered the amount of O2 to calculate the weight of nitric oxide obtained in the reaction).
Now, 160 g of O2 gives 120g of NO.
Therefore, 20 g of O2 gives `(120xx20)/160` g of N, 15 g of NO
Hence, a maximum of 15 g of nitric oxide can be obtained
The balanced equation for the reaction is:
`4NH_3(g) + 5O_2(g) -> 4NO(g) + 6H_2O(g)`
`4xx17 5xx3 4xx30`
= 68 g = 160 g = 120 g
Here 68 g of `NH_3` will `O_2` = 160g
:. 10 g of NH_3 will react will `O_2 = (160g)/(68g) xx 10g = 23.6 g`
But the amount of O2 which is actually available is 20.0 g which is less than the amount which is needed. Therefore, 02 is the limiting reagent and hence calculations must be based upon the amount of 02 taken and not on the amount of NH3 taken. From the equation,
160 g of 02 produce NO = 120 g
∴ 20 g of 02 will produce NO =120/160 x 20 = 15 g