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 874 mmHg and 33 ∘C is required to synthesize 18.0 mol of NO?

Answer 1

1) Write the balanced chemical equation

4NH3 + 5O2 ----> 4NO + 6H2O

2) Write the molar ratios

4 mol NH3 : 5 mol O2 : 4 mol NO : 6 mol H2O

3) Write the proportions with the desired quantity of NO and the unknown quantity of O2

5 mol O2 / 4 mol NO = x / 18.0 mol NO

Solve for x:       x = 18.0 mol NO * 5 mol O2 / 4 mol NO = 22.5 mol O2

4) Use the ideal gas equation to convert moles to volume

pV = nRT

V = nRT / p

n = 22.5 mol

R = 0.082 atm*liter / K * mol

T = 33 + 273.15 = 306.15 K

p = 874/760 atm = 1.15 atm

V = 22.5 mol * 0.082 atm*liter/K*mol * 306.15 K / 1.15 atm = 491.17 liter

Answer: 491 liters of O2