4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O ∆Hrxn = -906 kJ A)...

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O ∆Hrxn = -906 kJ A) Calculate the heat (in kJ) associated with the complete reaction of 0.127 Kg of NH3. B) What mass of O2 (g) is required to evolve 4605 kJ of energy

Solutions

Expert Solution

mass of NH3 = 0.127 kg = 127g

moles = mass / molar mass

= 127 / 17

= 7.47

moles of NH3 = 7.47

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O ∆Hrxn = -906 kJ

4 mol of NH3 gives --------------- -906 kJ

7.47 mol NH3 ----------------- ??

heat energy = 7.47 x - 906 / 4 = -1692 kJ

heat energy = - 1692 kJ

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O ∆Hrxn = -906 kJ

5 mol of O2 gives ------------------ ∆Hrxn = -906 kJ

?? mol O2 ---------------- 4605 kJ

moles of O2 = 25.41 moles

moles = mass / molar mass

25.41 = mass / 32

mass of O2 required = 813.2g

queen_honey_blossom answered 3 months ago

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