4 NH3 + 5 O2 �¨ 4 NO + 6 H2O

If the rate of disappearance of O2 is 0.772, what is the rate of appearance of NO?

0.772 x (4/5) = ??

To determine the rate of appearance of NO, we need to use the stoichiometric coefficients from the balanced chemical equation. The coefficients represent the ratios between the reactants and products.

From the balanced equation:
4 NH3 + 5 O2 -> 4 NO + 6 H2O

We can see that the stoichiometric coefficient of O2 is 5, and the stoichiometric coefficient of NO is also 4. This means that for every 5 moles of O2 that react, 4 moles of NO are produced.

The rate of disappearance of O2 is given as 0.772. This means that for every unit of time, 0.772 moles of O2 are consumed.

To find the rate of appearance of NO, we need to determine how many moles of NO are produced for every unit of time.

Using the stoichiometric ratio, we can set up the following proportion:

(0.772 moles O2 / 5 moles O2) = (x moles NO / 4 moles NO)

Solving for x, we can calculate the rate of appearance of NO:

x = (0.772 moles O2 / 5 moles O2) * 4 moles NO
x = 0.6176 moles NO

Therefore, the rate of appearance of NO is 0.6176 moles.