In a study to determine the rate of the following reaction:
2no(g)+o2(g) = 2no2(g)
the concentration of no was 0.0400 m at t = 5.0 s and 0.0225 m at t = 650.0 s. what is the average rate of the reaction during this time period?

The rate of reaction is - 2.868 * 10^-4 M/S

Explanation:

In the following reaction: 2 NO + O2 → 2NO2

NO will be consumed, when adding O2.

This means it will disappear. This is confirmed by the concentration at 5 seconds and 650 seconds.

A = -Δ[A] / ΔT   = [A2] - [A1] / T2 - T1

(0.225 - 0.04) / (650 - 5) = 0.185 / 645 = 2.868 * 10^-4 M/S

Since it will disappear, this number will be negative : - 2.868 * 10^-4 M/S

The rate of reaction is - 2.868 * 10^-4 M/S

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g