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 obtained starting only with `10.00 g` of ammonia and `20.00 g` of oxygen?

Answer

15g

Explanation

The balanced chemical equation for the given reaction is given as :
`4 NH_(3 (g)) + 5O_(2 (g)) to 4 NO_((g)) + 6H_(2)O_((g))`
`{:( 4 xx 17 g , 5 xx 32 g , 4 xx 30 g , 6 xx 18 g ), ( = 68 g , = 160 g , 120 g , = 108 g ):}`
Thus , 68 g of `NH_(3)` reacts with 160 g of `O_(2)`
Therefore , 10 g of `NH_(3)` reacts with `( 160 xx 10)/(68)` g of `O_(2)` or `23.53` g of `O_(2)`.
But the available amount of `O_(2)` is 20 g .
Therefore , `O_(2)` is the limiting reagent (we have considered the amount of `O_(2)` to calculate the weight of nitric oxide obtained in the reaction).
Now , 160 g of `O_(2)` gives 120 g of NO .
Therefore , 20 g of `O_(2)` gives `(120 xx 20)/(160)` g of N , or 15 g of NO .
Hence , a maximum of 15 g of nitric oxide can be obtained .