e Ideal Gas Law and Stoichiometry ) 100110 PartA The industrial production of ni

Question

e Ideal Gas Law and Stoichiometry ) 100110 PartA The industrial production of nitric acid (HNOs) is a multistep process. The first step is the oxidation of ammonia (NH3) over a catalyst with excess oxygen (02) to produce nitrogen monoxide (NO gas as shown by the unbalanced equation given here: what volume of O2 at 760. InmHg and 41°C is required to synthesize 21.0 mol of NO? Express your answer numerically in liters View Available Hint(s) TNI, (g)+7O2 (g)TNO(g)+7H2O(g) volume of O Submit Request Answer

Explanation / Answer

Ans. Balanced reaction:        4 NH3(g) + 5 O2(g) ---------> 4 NO(g) + 6 H2O(g)

According to the stoichiometry of balanced reaction, 4 mol NO is synthesized by 5 mol O2.

So,

Moles of O2 required to produce 21.0 mol NO = (5 mol O2 / 4 mol NO) x 21 mol NO

                                                            = 26.25 mol O2

# Given, Pressure, P = 760.0 mm Hg = 1.0 atm

            Temperature, T = 41.00C = 314.15 K

            Required moles of O2 = 26.25 mol (as calculated above)

Using Ideal gas equation:    PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol-1K-1

            T = absolute temperature (in K) = (0C + 273.15) K

Putting the values in above equation-

            1.0 atm x V = 26.25 mol x (0.0821 atm L mol-1K-1) x 314.15 K

            Or, V = 677.03 atm L / 1.0 atm

            Hence, V = 677.03 L

Therefore, required volume of O2 = 677.03 L

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