Question

given
NO + O3 --> NO2 + 02
what is the maximum volume of NO2 produced from 5.4mg NO and 9.6mg O3?

Answer 1

@Kystal how do i know which element to use to find the max?

Answer 2

You need to find which one is limiting reactant, either NO or O3

Answer 3

first you find the n of moles of NO and O3~

Answer 4

the from the equation we know that 1 mole of NO reacts with one mole of O3~
so the n of moles should be in the same amount~
but from the mass we know that one will be greater than another one~
the one with the less number of moles is referred as the limiting reagent~

Answer 5

Isn't this a limiting factor question

Answer 6

@.Sam. okay limiting reagant: NO gives us 1.8x10-4mol of NO2 and O3 gives us 2x10^-4mol NO2, so its the one with the least amount NO2 produced?

Answer 7

I mean limiting reagent

Answer 8

then NO is the limiting reagent

Answer 9

@Kystal so our limiting reagent is NO

Answer 10

if you did your calculations correctly then of course it is, limiting reagent means that it is the reactant that limits the amount of compound that can be made

Answer 11

yeap~ NO is the limiting reagent~

Answer 12

@bronzegoddess if your mole values are correct then yes

Answer 13

okay, thank you.. for max values look for limiting reagent and for minimum as well?

Answer 14

The limiting reagent can be noted as the compound you need the most of to make the product

Answer 15

what ever makes the least amount of product will be the limiting reagent.

Answer 16

\[1.8 \times 10^{-4} \text{mol of NO2}\times \frac{22.4 ~dm^{-3}~of NO_2}{1mol~of~NO_2}\]

Answer 17

so now from the equation~
we know that n(NO)=n(NO2)
=1.799x10^-4 mol
n(NO2)=V/22.41 (using the formula n=V/22.41 at SPT)
so V(NO2)=1.799x10^-4 (22.41)
=4.03x10^-3 L (or 4.03 mL)