If this is REALLY 906 kJ/MOLE that is 906 kJ/17 g NH3.
Then 17 x (500/906) = ? g NH3 required.
What mass of ammonia is necessary to produce 500 kJ of heat?
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) ΔrH = -906 kJ mol-1
1 answer
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) ΔrH = -906 kJ mol-1
1 answer