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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

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

PV=nRT

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