Question

The industrial production of nitric acid (HNO3) is a multistep process. The first step is the oxidation of ammonia (NH3) over a catalyst with excess oxygen (O2) to produce nitrogen monoxide (NO)gas as shown by the unbalanced equation given here:

?NH3(g)+?O2(g)→?NO(g)+?H2O(g)

What volume of O2 at 988 mmHg and 37 ∘C is required to synthesize 18.5 mol of NO?

Answer 1

    4NH3(g)+5O2(g)→4NO(g)+6H2O(g)

4 moles of NO produced from 5 moles of O2

18.5 moles of NO produced from = 5*18.5/4   = 23.125 moles of O2

PV = nRT

P   = 988mmHg = 988/760 = 1.3 atm

R   = 0.0821L-atm/mole-K

T = 37C⁰ = 37 + 273 = 310K

V = nRT/P

= 23.125*0.0821*310/1.3   = 452.73L

4 moles of NO produced from 6 moles of H2O

18.5 moles of NO produced from = 6*18.5/4   = 27.75moles of H2O

PV = nRT

P   = 988mmHg = 988/760 = 1.3 atm

R   = 0.0821L-atm/mole-K

T = 37C⁰ = 37 + 273 = 310K

V = nRT/P

= 27.75*0.0821*310/1.3   = 543.28L