Question

Just want to make sure I'm on the right track.

NI3 + 2O -> NO2 + 3I and 0.206 mols of oxygen react every minute in a 750.0mL vessel.
a. rate of oxygen consumption = 0.275 M/min ???
I also have to find rate of NI3 consumption and the rate of nitrogen dioxide and iodine production.

Answer 1

The oxygen in the reaction is atomic oxygen O or molecular oxygen O2?

I will said that if the oxygen react 0.206 moles and the container is 750mL you have less than a liter then that is not the concentration in Molarity,

1L x 0.206 mol/0.750 L= ????

Then: the rate of the reaction is

Rate= -Δ[O]/2Δt = -Δ[NI3]/Δt =Δ[NO2]/Δt = Δ[I]/3Δt

so you will have to divide the rate of consumption of oxygen by 2 to find the rate of consumption of NI3 and the rate of production of NO2.

for the rate of production of I, you will have to multiply by 3 and divide by 2 the rate of oxygen