Question

PART B ONLY<

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 912 mmHg and 33 ∘C is required to synthesize 16.0 mol of NO?

volume of O2 = 419 L

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. volume of H2O =

Answer 1

Balance the equation:

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

A) As per the balanced stoichiometric equation, the molar ratios of O₂ and NO is 5:4, i.e, 5 moles of O₂ are required to synthesize 4 moles of NO. We need to find out the moles of O₂ required from the given data by using the ideal gas law.

Therefore, moles of O₂ required to produce 16.0 mol NO = (16.0 mol NO)*(5 mol O₂/4 mol NO) = 20.0 mol

The ideal gas law gives PV = nRT where P = pressure of the gas in atmospheres, V = volume of the gas in L; T = temperature of the gas in Kelvin scale and n = number of moles of the gas; R = gas constant having a value of 0.082 L-atm/mol.K.

Find out P now: We know that 760 mmHg = 1 atm; therefore, P = 912 mmHg = (912 mmHg)*(1 atm/760 mmHg) = 1.2 atm; n = 20.0 mol and T = (33 + 273) K = 306 K. Plug in values in the expression to obtain

(1.2 atm)*V = (20.0 mol)*(0.082 L-atm/mol.K)*(306 K)

===> V = (20.0 mol)*(0.082 L-atm/mol.K)*(306 K)/(1.2 atm) = 418.2 L

The volume of O₂ required = 418.2 (ans).

B) Again, refer to the balanced stoichiometric equation. The molar ratio of NO and H₂O is 4:6, i.e, 6 moles of H₂O are produced for 4 moles of NO. The reaction conditions remain the same which means that the pressure remains at 1.2 atm and the temperature remains at 306 K.

Moles of H₂O produced for 16.0 mol NO production = (16.0 mol NO)*(6 mol H₂O/4 mol NO) = 24.0 mol H₂O

Use the equation of state with n = 24.0 mol and obtain

(1.2 atm)*V₁ = (24.0 mol)*(0.082 L-atm/mol.K)*(306 K)

===> V₁ = (24.0 mol)*(0.082 L-atm/mol.K)*(306 K)/(1.2 atm) = 501.84 L

The volume of H₂O produced under the same conditions = 501.84 L ≈ 502 L (3 significant digits) (ans)