Question

2NO (g) + O2 (g) --> 2NO2 (g)
In one experiment, 0.891 mol of NO is mixed with 0.517 mol of O2. Determine which of the two reactants is the limiting reactant. Calculate also the number of moles of NO2 produced.

Answer 1

@Rushwr

Answer 2

Okai look at the reaction ratio of NO: O2
That is 2:1 right?
That means 2 moles of NO reacts with 1 mole of O2.
SO we know that we have 0.517 moles of O2 , according to the ratio we should have twice the no. of moles of NO.
So to completely react with the O2 present we need 1.034 moles of NO. But we only have 0.891moles of NO therefore we can see NO is the limiting agent.

Answer 3

Okay so how do you do the NO2 part?

Answer 4

So we have to relate it with the limiting agent.
According to the equation if we react 2 moles of NO then 2 moles of NO2 has to be produced.
Therefore we can say the no. of moles of NO2 produced by 0.891 moles NO ois 0.891 moles
Did u get it ?

Answer 5

Sorta could you set it up in an equation. I think that would help

Answer 6

Cause I get mixed up what goes with what.

Answer 7

Look at the equation

Answer 8

we figured out that the NO is the limiting agent right? That means at the end of the reaction no NO present.
Since the no. of moles of NO = no. of moles of NO2
We can say no. of moles of NO2 = 0.891

Answer 9

Right

Answer 10

Okay so that's what you use to multiple with since it's the limiting reactant?

Answer 11

yeah

Answer 12

Okay so you take the
0.891 * (2 NO/ 1 O2) =
.0.891 * (1O2 / 2 NO) =

Answer 13

no no. u don't have to do that.
The no. of moles of NO2 straight away equals to the no. of moles of NO reacted

Answer 14

Okay so the NO2 produced would be 0.891?

Answer 15

yes

Answer 16

Alright. Thank you!

Answer 17

NO problems :)

Answer 18

problem *