Limiting reagent! Number of Moles?

Question

For example if i have 2 C2H6 + 7 O2 = 4 CO2 + 6 H2O, and i have 100g of C2H6 and 100g of 02, when finding the number of moles of each, do i have to times the Mass divided by Molar Mass by the coefficient? Or do i times the molar mass by the coefficient before dividing? Or do i not have to times at all? What im trying to say is HOW DO I FIND THE NUMBER OF MOLES?? Please read this question carefully (im confused enough as it is) and answer asap!!

Cheers all!

Yes, well actually let me rephrase. I need to find the number of moles in order to find the limiting reagent (but its ok, i understand limiting reagents) Thankyou for ur answer (im still in the dark though)

Answer 1

dear i understood ur confusion
if ur aim or question is find no of moles of C2H6 in 100gm --divide by molar mass only not coefficient)
if ur are dealing with question pertaining to eqn only divide 100 by molar mass by the coefficient
also remember 2 moles of C2H6 react with 7 moles of O2 ie in 100 gm of each one will be in excess...

Answer 2

If you have a bag of tennis balls, and you want to know how many tennis balls you have, you divide the total weight of the bag by the weight of one ball.

If you have a sample of ethane, and you want to know how many moles of ethane you have, you divide the weight of the sample by the weight of 1 mole.

That's all there is to it. Don't let anyone confuse you.

The coefficient only comes in at the next stage. The balanced equation tells you that you need 7 mol oxygen for complete combustion of 2 mol ethane.

I think your problem may be that you are trying to do two different steps at once.

In this problem, you have roughly 3 mol each of ethane and oxygen (I leave it to you to work out the exact numbers, the way I just told you). If you have (as you do have) less than 7 moles oxygen for every 2 mol ethane, ethane is present in excess and oxygen is the limiting reagent.

Answer 3

moles = mass / molecular weight...

molecular weight = sum of (atomic weights x number of atoms) of the elements in the molecule.....

for example....

100 g C2H6....

molecular weight C2H6 = 2 x 12 + 6 x 1 = 30 g/mole
the 2 is because you have 2 carbons, the 12 is the atomic mass of carbon, the 6 is 6 hydrogens, the 1 is the atomic mass of hydrogen.

so number of moles = 100 g / (30g/mole) = 3.33 moles

*****

100 g O2

molecular weight = 2 x 16 = 32 g/mole

moles = 100 g / (32 g/mole) = 3.13 moles

*********
as to limiting reagent, from your balanced equation,,,

2 C2H6 + 7 O2 ---------> 4 CO2 + 6 H2O

2 moles of C2H6 require 7 moles O2

so, 3.33 moles C2H6 would require 7/2 x 3.33 moles of O2 = 11.7 moles O2..

since you only have 3.13 moles O2, O2 is the limiting reagent.

Answer 4

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Answer 5

2Al +CuCl2 > 2AlCl3 +3Cu discover mols. 5.0g Al X 1mol/26.98g = 0.1853mol Al 5.0g CuCl2 X 1mol/134.45g = 0.03719mol CuCl2 discover limiting reactant. i think of CuCl2 0.03719mol CuCl2 X 2mol Al/3mol CuCl2 = 0.02479mol Al you have plenty greater Al than that, so CuCl2 limits and willpersistent reaction. 0.03719mol CuCl2 X 3mol Cu/3mol CuCl2 sixty 3.55g/1mol = 2.363 grams of copper produced. Now, you're in a position to do 2.