In: Chemistry
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
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
B)
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