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?

Explanation

The rection involved in the manufacturing process is :
`4NH_(3)(g)+50_(2)(g)rarr4NO(g)+6H_(2)O(g)`
`4xx17=68g 5xx32=160g 4xx30=120g`
Form the available data:
`68g of NH_(3)` will react with `O_(2) = 160g`
`10 g of NH_(3)` will react with `O_(2) = (160g)/(68g)xx(10g)=23.6 g` But oxygen which is actually available (20.0g) is less than the amount which is needed.Therefore oxygen limiting reactant.
Now, 160 g of `O_(2)` will form NO =120 g
`therefore 20 g of O_(2)` will from `NO = (120 g)/(160 g)xx(20g)=15g `