Gaseous Reactions

20 mols gaseous ammonia and 15 mols oxygen gas react to form nitrogen monoxide gas and water vapor.

NH_3(g) + 5O_2(g)  -> 4NO(g) + 6 H_2O(g)

If the reaction occurs at 30\deg C and atmospheric conditions of pressure what will be the final volume of the unreacted component and products.

Step 1: Determine the limiting reactant

  • Assuming all reactants react to determine which reactant produces less product

20 mol NH_3 \cdot \frac{4 mol NO}{4 mol NH_3}=20 mols NO produced

15 mol O_2 \cdot \frac{4 mol NO}{5 mol O_2}=12 mols NO produced

Step 2: Determine # of mols present after reaction

15 mol O_2 \cdot \frac{6 H_2O}{5mol O_2} -> 18 mol H_2 O produced

15 mol O_2 \cdot \frac{4mol NH_3}{5 mol O_2} -> 12 mols NH_3 react

20 mol NH_3 -12mol NH_3 = 8mol NH_3

8mol NH_3 +18 mol H_2O + 12 mols NO = 38 mols

Step 3: Ideal gas law Assumption

Assume water vapor, NH_3 and NO behave as IG


  • Solve for Volume
    • V=\frac{nRT}{p}
  • Substitution
    • \frac{(38mols)(0.08206 \frac{L atm}{mol K})(273.15+30 K)}{1 atm}=945.3 L