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 684 mmHg and 27 ∘C is required to synthesize 20.0 mol of NO?
The balanced equation is 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
According to the balanced equation ,
4 moles of NO produced from 5 moles of O2
20.0 mol of NO produced from M moles of O2
M = ( 20.0 x 5 ) / 4
= 25.0 mol
Calculation of volume of O2 :-
We know that PV = nRT
Where
P = pressure = 684 mm Hg = 684 / 760 atm Since 1 atm = 760 mm Hg
= 0.9 atm
V = volume = ?
N = number of moles = 25.0 mol
R = gas constant = 0.0821 L-atm / (mol-K)
T = temperature = 27 oC = 27+273 K = 300 K
Plug the values we get
V = nRT / P
= ( 25.0 x0.0821x300) / 0.9
= 684.2 L
Therefore the volume of oxygen required is 684.2 L
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