Question

Nitric acid is a key industrial chemical, largely used to make fertilizers and explosives. The first step in its synthesis is the oxidation of ammonia. In this reaction, gaseous ammonia reacts with dioxygen gas to produce nitrogen monoxide gas and water.

Suppose a chemical engineer studying a new catalyst for the oxidation of ammonia reaction finds that 645. liters per second of dioxygen are consumed when the reaction is run at 195.oC and 0.88 atm. Calculate the rate at which nitrogen monoxide is being produced. Give your answer in kilograms per second. Be sure your answer has the correct number of significant digits.

Answer 1

Answer: Rate at which nitrogen monoxide is being produced 0.355Kg /s

Step-by-step explanation:

• First we review the equation that represents the oxidation process of the NH3 to NO.

4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)

• Second we gather the information what we are going to use in our calculations.

O2 Volume Rate = 645 L /s

Pressure = 0.88 atm

Temperature = 195°C + 273 = 468K

NO molecular weight = 30.01 g/mol

• Third, in order to calculate the amount of NO moles produced by 645L/ s of O2, we must find out, how many moles (n) are 645L O2 by using the general gas equation PV =n RT

Let´s keep in mind that using this equation our constant R is
`(0.08205 Lxatm)/(Kxmol)`

PV =n RT

n= PV / RT

n= [ 0.88atm x 645L/s] / [ (0.08205 Lxatm/Kxmol) x 468K]

n= 14.781 moles /s of O2

• Fourth, now by knowing the amount of moles of O2, we can use the equation to calculate how many moles of NO will be produced and then with the molecular weight, we will finally know the total mass per second .

`(14.781moles O2)/(s) x  (4moles NO)/(5 moles O2) x(30.01gNO)/(1 mol NO) x(1Kg NO)/(1000gNO) = <strong>0.355Kg/s`