4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l). Suppose that in this reaction water is being formed at a rate of 28.9 moles liter-1 sec-1. At what rate (in moles liter-1 sec-1) is ammonia used up?
To determine the rate at which ammonia (NH3) is being used up in the reaction, we need to use the stoichiometry of the balanced equation.
The balanced equation is: 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l)
From the balanced equation, we can see that 4 moles of NH3 react to produce 6 moles of H2O. This means that the stoichiometric ratio between NH3 and H2O is 4:6, or simplified 2:3.
Since we are given the rate at which water is being formed (28.9 moles liter-1 sec-1), we can use the stoichiometric ratio to determine the rate at which NH3 is being used up.
Using the ratio 2:3, we can calculate the rate of NH3 consumption as follows:
Rate of NH3 consumption = (Rate of H2O formation) x (stoichiometric ratio)
= 28.9 moles liter-1 sec-1 x (2/3)
≈ 19.3 moles liter-1 sec-1
Therefore, the rate at which ammonia is being used up in this reaction is approximately 19.3 moles liter-1 sec-1.