Ammonia reacts with oxygen according to the equation 4nh3(g)+5o2(g)→4no(g)+6h2o(g),δhrxn =−906 kj calculate the heat (in kj) associated with the complete reaction of 155 g of nh3. express the heat in kilojoules to three significant figures.

2.06 × 10 ³ kJ = 2,060 kJ

Explanation:

1) Balanced chemical equation (given)

4NH₃ (g) + 5O₂ (g) → 4NO (g) + 6H₂O (g), δH rxn = − 906 kJ

2) Calculate the number of NH₃ moles in 155 g

i) molar mass of NH₃ = 14.007 g/mol + 3×1.008 g/mol = 17.031 g/mol

ii) number of moles = mass in grams / molar mass = 155 g / 17.031 g/mol = 9.10 mol of NH₃

3) Caculate the energy using proportions with the mole /energy ratio from the balanced chemical equation:

4 moles NH₃ / (-906 kJ) = 9.10 mol NH₃/ x

⇒ x = 9.10 mol NH₃ × (-906 kJ) / 4 mol NH₃ = - 2,061.15

Which rounded to three significan figures is 2.06 × 10 ³ kJ = 2,060 kJ

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