8.25 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?

Answers (1)

Answer-

we have, 

number of moles(n) = given mass/ molecular mass   ------------------------------(eq.1)

No. of moles of ammonia(NH_{3})  = 10/17 = 0.588

No. of moles of oxygen (O_{2}) = 20/32= 0.625

Balanced Reaction   4NH_{3}+5O_{2}\rightarrow 4NO + 6H_{2}O

Here we see that 4 moles of ammonia required 5 moles of oxygen. So

0.588 moles of ammonia = (5/4)*0.588 = 0.735 moles of  (O_{2}). But we have only 0.625 moles of (O_{2}).

It means oxygen is a limiting reagent and the maximum weight of nitric oxide(NO) can be produced by 0.635 moles of (O_{2}) 

So, 5 moles of (O_{2}) produced 4 moles of C.

therefore 0.625 moles of (O_{2}) = (4/5)*0.625 = 0.5 moles of (NO).

from Eq. 1 

mass of (NO) = number of moles(NO) * molecular weight(NO)

          = 0.5* 30

          = 15 g

Alternate Method

directly consider the molecular weight

(17*4) g of NH3 required (5*32) g of O to produce (30*4) g of NO

So, 10g of NH3 required=  (5*32/17*4)*10 = 23.5g of O. But we have only 20g (means O is limiting reagent) whatever the max. NO produce is from 20g of O.

and we know that 5*32g of O produce 30*4 g of NO 

So, 20g of O produce =(30*4/5*32)*20 g of NO = 15g of NO

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