Question

The Ideal Gas Law and Stoichiometry

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)

Part A

What volume of O2 at 836 mmHg and 27 ∘C is required to synthesize 10.5 mol of NO?

Part B

What volume of H2O(g) is produced by the reaction under the same conditions?

Express your answer to three significant figures and include the appropriate units.

Answer 1

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

Part A

4 mol NH3 = 5 mol o2 = 4 mol NO = 6 mol H2O

10.5 mol NO reacts = 10.5*5/4 = 13.125 mol O2

volume of o2 = nRT/P

             = 13.125*0.0821*300.15/(836/760)

             = 294 L

part B

   4 mol NO = 6 mol H2O

no of mol of water produced = 6*10.5/4 = 15.75 mol

volume of water produced = nRT/P

             = 15.75*0.0821*300.15/(836/760)

             = 353 L