Nitric acid, a major industrial and laboratory acid, is produced commercially by the multi-step Ostwald process, which begins with the oxidation of ammonia:

Question

Step 1:  NH3(g) + O2(g)   NO(g) + H2O(l)

Step 2: NO(g) + O2(g)  NO2(g)

Step 3: NO2(g) + H2O(l)  HNO3(l) + NO(g)

Assuming 100% yield in each step, what mass (in kg) of ammonia must be used to produce 7.805 kg of nitric acid.  Enter to 3 decimal places.  Are the equations balanced?

Answer 1

First thing to do is to balance each reaction in steps 1, 2 and 3.

Step 1: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)

Step 2: 2NO(g) + O2(g)  2NO2(g)

Step 3: 3NO2(g) + H2O(l)  2HNO3(l) + NO(g)

Looks like if you start with 4 moles NH3 you end up with 2 moles HNO3. Converting kg to g to moles and then moles to g to kg, we have...

7.805 kg HNO3 x 1000 g/kg = 7805 g

7805 g HNO3 x 1 mol HNO3/63.01 g = 123.9 moles HNO3

123.9 moles HNO3 x 4 moles NH3 / 2 moles HNO3 = 61.95 moles NH3 needed

61.95 moles NH3 x 17 g NH3/mol = 1053 g NH3

1053 g NH3 x 1 kg/1000 g = 1.053 kg NH3 needed