# LIMITING reactions

1. Mar 2, 2005

### megas

LIMITING reactions(plz help me!!! (is it that hard?)NEED A CHALLENGE?

hi, i was wondering if anyone could help me out with my chemistry 1 stuff. The proffessor isnt explaining it very well to me. so....well, heres the question, i need help with limiting reactions, O2 + 2No2-->2NO2, if they gave me 10.00g of 02, and 20g of 2NO2 what is the amount of 2NO2, the limiting reaction and amount left? i could use the answere, but i would really like to know how to do it thanks

Last edited: Mar 2, 2005
2. Mar 3, 2005

### chem_tr

The main problem is that you will have to find which reactant is excess, so the other will be the limiting reagent. To find this, you'll use mole calculations. Which one is excess, 10.00 g of O2 or 20 g of NO2? You can use O=16 and N=14 g/mole.

3. Mar 3, 2005

### Gokul43201

Staff Emeritus
Make sure the equation you provide is correct. In this case it is not. You have NO2 as a reactant and product.

4. Mar 3, 2005

### megas

sorry

the right formula is O2 + No2 -> 2No2,.. ok thanks i get it now, lol, (dragonball Z rules)

5. Mar 3, 2005

### chem_tr

No, this is incorrect either. Please take a good look at your reaction; on the left, you have three oxygens and one nitrogen, but on the right, you have two nitrogens and four oxygens.

I bet the reaction should be like this:

$$O_2 + 2NO \longrightarrow 2NO_2$$

6. Mar 3, 2005

### megas

sorry chem_tr is right!!! (allways right) well, im sorry for the mix up, my first time posting. thanks for catching that chem!

7. Mar 3, 2005

### ShawnD

I first learned to do these by looking for how much of B you would need to react with the given amount of A, then decide which of the reactants is in excess then do the calculation through. I later found out that it was a lot easier and faster to just do the entire calculation through for each of the reactants separately then use whichever value gives the lowest yield. Takes a lot less thinking so it's less prone to errors.

8. Mar 4, 2005

### chem_tr

Yeah, this is a nice approach indeed. Thank you for reminding this.

9. Mar 4, 2005

### megas

ok thanks so much for your help guys! until next time =)