Question

in ostwald process for the manufacture of nitric acid,the first step in was 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 gram of ammonia and 20 gram of oxygen ?​

Answer 1

Answer:

Firstly,

 The balanced equation for the process is

$4NH_{3} + 5O_{2} ---> 4NO_{2} + 6H_{2}O$

$4NH_{3}   = 17 *  4 = 68 g\\\\5O_{2} = 32 *5 = 160g\\\\4NO_{2} = 30* 4 = 120g$

According to the equation,

68g of ammonia require 160g of the $O_{2}$ for oxidation.

Therefore , 10g of$NH_{3}$ would require

= $\frac{160}{68}$ * 10

=  23.53g of $O_{2}$.

Since , the used $O_{2}$ is only 20g,

$O_{2}$ is the limiting reagent.

thus,

 160g of  $O_{2}$ gives NO = 120g

20g of  $O_{2}$ will give NO = $\frac{120}{160} * 20$

                                                          = 15.00g