Question

When 25.0 grams of ammonia are reacted with 37.0 grams of oxygen in a closed, expandable container kept at STP, the reaction gives a 48% yield. What final volume of gas is in the container?

4 NH₃ (g) + 3 O₂ (g) = 2 N₂ (g) + 6 H₂O (l)​

Answer 1

Moles of NH3 = 25g / 17g/mol = 1.47 moles

Moles of O2 = 37g / 32 = 1.156 moles

Limiting reagent is NH3

Because 1 mol of NH3 consumes 0.75 mol of O2

Theoretical yield of N2

= 2 mol of N2 x 1.47 mol of NH3 / 4 mol of NH3

= 0.735 mol of N2

Mass of N2 = 0.735 mol x 28g/mol = 20.58 g

% yield = actual yield / theoretical yield

Actual yield = 0.48*20.58 = 9.8784 g of N2

Moles of N2 formed = 9.8784 g / 28g/mol = 0.3528 mol

Moles of O2 unreacted = 1.156 - 1.5*0.3528 = 0.6268

Moles of NH3 unreacted = 1.47 - 2*0.3528 = 0.7644

Total moles of gas in the container = 1.745 moles

Final volume of gas at STP = nRT/P

= 1.745 mol x 8.314 J/mol·K x 273.15 K / 101325 Pa

= 0.0391 m3 = 39.1 L