Question

determine the amount of heat (in kj) associated with the production of 9.99 × 10⁴ g of ammonia according to the following equation.
n₂(g) + 3h₂(g) → 2nh₃(g)
δh°rxn = -92.6 kj/mol
assume that the reaction takes place under standard - state conditions at 25.0 °c. round your answer to 3 significant digits.

Explanation:

Step1: Calculate moles of NH₃

Molar mass of NH₃ = 14.01 + 3×1.008 = 17.034 g/mol
Moles of NH₃ = $\frac{9.99×10^4\ \text{g}}{17.034\ \text{g/mol}} ≈ 5864.7\ \text{mol}$

Step2: Relate moles to ΔH°

Reaction produces 2 mol NH₃ with ΔH° = -92.6 kJ
Heat for 5864.7 mol NH₃ = $5864.7\ \text{mol} × \frac{-92.6\ \text{kJ}}{2\ \text{mol}} ≈ -2.73×10^5\ \text{kJ}$

Answer:

$-2.73×10^5$