Limiting Reagent question

[Lizzie Bartlet]

It's simple guys. Don't laugh at me.

What is the limiting reagent when 34 grams of ammonia form 26 grams of nitrogen in a reaction that runs to completion?

2NH3 + 3CuO ----> N2 + 3Cu + 3H2O

The answer is CuO.

Can someone please work me through the math? I know it's something simple that I'm just not getting.

I have that the molecular weight of NH3 is 17 grams and there are two moles of it for 34 grams. The molecular weight of N2 is 28 grams. But how do you know the 34 grams was enough of NH3 to make N2? I just can't connect the dots.

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[marbou3]

This is how I see it... if theoretically all of the ammonia was reaction (34g or 2mols) then 28 g of N2 would be formed. In the problem though it tells you that 26 g were formed. You can concluded that the CuO is the limiting reagent.

34 g * (1mol/ 17g) * (1 mol N2/ 2 mol NH3) * (28 g/ 1 mol N2)= 28 g of N2 that could be produced with 34 g of NH3

......so if NH3 is not limiting the reaction that only leaves one possibility. I hope this helps!

 

[Kaustikos]

This is how I see it... if theoretically all of the ammonia was reaction (34g or 2mols) then 28 g of N2 would be formed. In the problem though it tells you that 26 g were formed. You can concluded that the CuO is the limiting reagent.

34 g * (1mol/ 17g) * (1 mol N2/ 2 mol NH3) * (28 g/ 1 mol N2)= 28 g of N2 that could be produced with 34 g of NH3

......so if NH3 is not limiting the reaction that only leaves one possibility. I hope this helps!

That's how I figured it out. There are only two reagants, and you have more than enough of the first reagant, but still come out short for products.

 

[MLT2MT2DO]

This is how I see it... if theoretically all of the ammonia was reaction (34g or 2mols) then 28 g of N2 would be formed. In the problem though it tells you that 26 g were formed. You can concluded that the CuO is the limiting reagent.

34 g * (1mol/ 17g) * (1 mol N2/ 2 mol NH3) * (28 g/ 1 mol N2)= 28 g of N2 that could be produced with 34 g of NH3

......so if NH3 is not limiting the reaction that only leaves one possibility. I hope this helps!

I'm confused, if you do the same with 3CuO you get the same answer...maybe I did my math wrong

238.95*(1/79.65)(1/3)(28/1)= 28g N2 as well...

 

[Arsenic]

I'm confused, if you do the same with 3CuO you get the same answer...maybe I did my math wrong

238.95*(1/79.65)(1/3)(28/1)= 28g N2 as well...

where are you getting the 238.95?

i reasoned it all out the same way the others did. if you had used up all the 34 g of NH3, thats 2 moles of NH3 and since the stoichiometry is 2:1 between NH3 and N2, you would have gotten 1 full mole of N2 for these 2 moles of NH3. You didn't get your 1 mole of N2 though, you only got 26 g of N2 which is 0.93 mole, so just by that the reaction must to have been limited by the other reagent... CuO.

You can figure out how much CuO was consumed by looking at the stoichiometry too if you wanted, since its a 3:1 ratio of CuO to N2, and you ended up with 0.93 mole of N2 (26 g) you must have used 0.31 mole of CuO which would have been about 24.6 g of CuO.

 

[MLT2MT2DO]

where are you getting the 238.95?


You can figure out how much CuO was consumed by looking at the stoichiometry too if you wanted, since its a 3:1 ratio of CuO to N2, and you ended up with 0.93 mole of N2 (26 g) you must have used 0.31 mole of CuO which would have been about 24.6 g of CuO.

238.95=3 Moles of CuO? Maybe I'm just overthinking this or assuming, I should probably go review inorganic chem.

 

[Arsenic]

238.95=3 Moles of CuO? Maybe I'm just overthinking this or assuming, I should probably go review inorganic chem.

haha, no. i see what you were saying. i actually made a mistake in that 2nd part, you would have needed 3 times as many moles of CuO for that reaction so if you ended up with 0.93 moles of N2 there must have been 2.79 moles of CuO consumed. sorry about that, i was a chem mjr but havent done a stoichiometry problem since 2004. think of it this way, had there been 3 full moles of CuO like you put (238.95 g) you would been able to use all the NH3 and the rxn wouldn't have been limited at all... you would have gotten 1 full mole of N2. but since you ended up with just 0.93 moles of N2 then you must have had 2.79 moles of the CuO, so you know its the one that limited you since you had 2 full moles of the NH3 which was more than you needed to cover the 0.93 x 2 = 1.86 moles of NH3 that was consumed to make the 0.93 moles of N2. had it been 3 moles of CuO the problem wouldnt have been limiting.

 

[marbou3]

I'm confused, if you do the same with 3CuO you get the same answer...maybe I did my math wrong

238.95*(1/79.65)(1/3)(28/1)= 28g N2 as well...

Nowhere in the problem does it tell you the amount of CuO. In theory if you did have 238.95g of CuO then there would be no limiting reagent because they would both run out at the exact same time. The coefficients only tell you the stoichiometric ratios between the different elements. You could figure out how much CuO you used by multiplying the (moles of NH3)*(molar ration of CuO/NH3)*(molar mass of CuO).

 

[MLT2MT2DO]

Ah ok, gotcha, ty both of you. My problem was I was assuming since 34g=2NH3 in the equation I could go ahead and do the same thing with the 3 moles of CuO. Hence why I couldn't come up with a limiting reactant.

Sorry about the thread hijack liz