In: Chemistry
What volume of O2 at 836 mmHg and 31 ∘C is required to synthesize 12.5 mol of NO?
Additional info: 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)
Here is how we are going to calculate volume of O2 required to obtain mols of NO:
Mols of NO ---> mols of O2 ----> Volume of O2
4 NH3 (g) + 5 O2 (g) ----------> 4 NO (g) + 6 H2O (g)
12.5 mol NO (5 mol O2 / 4 mol NO) = 15.625 mols of O2 necessary to obtain 12.5 mols of NO.
PV = n R T
P = pressure (using atmospheres)
n =number of mols of substance
V= volume in Liters
R= is a constant = 0.0821 L . atm . K-1 . mol-1
T= temperature (in Kelvins)
V = n R T / V
836 mmHg (1 atm / 760 mmHg) = 1.10 atm
K = 273 + 31 = 304 K
V = (15.625 mols) x (0.0821 L . atm . K-1 . mol-1) x (304 K) / 1.10 atm
V = 354.52 L of O2