Check if im right please:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) At a specific time in the reaction, ammonia is disappearing at rate of 0.068 mol/(L • s). What is the corresponding rate of production of water?
I did
Rate of Production H20 = 1/6(0.068)
= 0.0113 mol/(L*s)
Am i right?>
I would have done it this way/.
0.068 x *6 mols H2O x 4 mols NH3) = 0.068 x 6/4 = ?
Yes, you are correct!
To determine the rate of production of water, you need to consider the stoichiometric ratio between ammonia (NH3) and water (H2O) in the balanced chemical equation.
According to the balanced equation: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g), the ratio of NH3 to H2O is 4:6 (or 2:3).
Given that the rate of disappearance of NH3 is 0.068 mol/(L • s), you can find the corresponding rate of production of water by using the stoichiometric ratio:
Rate of Production H2O = (2/3) * Rate of Disappearance NH3
= (2/3) * 0.068 mol/(L • s)
= 0.0453 mol/(L • s)
So the corresponding rate of production of water is 0.0453 mol/(L • s), not 0.0113 mol/(L • s).