In Ostwal'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.

Explanation

Chemical reaction is
`4NH_(3)+5O_(2)to 4NO+6H_(2)O`
`4" moles"NH_(3)-5" moles of "O_(2)`
Given 10 gms of `NH_(3)`
`:.`No. Of moles= `(10)/(17)=0.588 " moles"`
Given 20 gms of `NH_(3)`
`:.` No. Of moles = `20/32 = 5/8 =0.4`
4 moles of ammonia reacts with 5 moles of `O_(2)` 0.588 moles of `NH_(3) ......?`
`(0.588)/(4)xx5=0.735 " moles" `
5 moles of `O_(2)` reacts with 4 moles of `NH_(3)`
0.4 moles of `O_(2)` ...... ?
`(0.4)/(5)xx4 = (1.6)/(5)=0.32`
`to` Here `O_(2)` is not present in sufficient amount `NH_(3)` has suffivient in amount
5 moles of `O_(2)to ` 4 moles of NO.
5x x 32 gms of `O_(2) to `4 xx 30 gms of No.
20 gms of `O_(2) to `?
`(20 xx 4 xx 30)/(5 xx 32)=(20 xx 24)/(32)=(120)/(8)=15 ` gms