Question

In: Chemistry

What volume of O2 at 836 mmHg and 31 ∘C is required to synthesize 12.5 mol...

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)

Solutions

Expert Solution

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

  • First, we balance the equation:

4 NH3 (g) + 5 O2 (g) ----------> 4 NO (g) + 6 H2O (g)

  • Then, we calculate mols of O2 that are necesary to obtain 12.5 moles of NO:

12.5 mol NO (5 mol O2 / 4 mol NO) = 15.625 mols of O2 necessary to obtain 12.5 mols of NO.

  • Now, we need to calculate the Volume of O2 required to obtain 12.5 mols of NO. We are going to do this using the Ideal Gas Law Equation:

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)

  • We rearrange the equation to obtain volume:

V = n R T / V

  • Now, we convert mmHg into atmospheres. 1 atm = 760 mmHg

836 mmHg (1 atm / 760 mmHg) = 1.10 atm

  • We convert °C into Kelvins.

K = 273 + 31 = 304 K

  • Now, we solve for Volume of O2, we substitute with values into the equation, since we have mols of O2 (calculated before: 15.625 mols), temperature (304 K) and pressure (1.10 atm):

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


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