Calculate % Yield: Pb(NO3)2 + LiBr -> PbBr2

[parwana]

For the reaction Pb(NO3)2 (aq) + LiBr PbBr2 precipitates out.
Calculate the percent yield (%) if 8.992 g of PbBr2 precipitates when 1.350 L of 0.0550 M Pb(NO3)2 (aq) and 1.400 L of 0.0408 M LiBr (aq) are mixed.


This is ridiculously confusing, please help with this. Why does it give the L and M of both equations, please teach me how to do this in steps.

 

[Sirus]

You are looking for percent yield, which is equal to [(experimental yield)/(predicted yield)]x100. In this problem, you need to find predicted yield using stoichiometry (unit analysis). Experimental yield is given to you as 8.992 g. See this thread to get you started: https://www.physicsforums.com/showthread.php?t=49285&page=2&pp=15. I made some mistakes at the beginning so go to post #16 for the corrected unit analysis. Can you try to do unit analysis here to get the mass of PbBr_{2}?

They give you L and M because M=mol/L (meaning you can figure out the moles of each reactant). Once you have done that, use unit analysis and the balanced equation to find the moles of PbBr_{2}, and thus the mass.

 

[Sirus]

Hmm...forgot one thing. There may also be a limiting reagent. Find the moles you have of each reactant as indicated above, then check to see if these two amounts are in the same ratio as the compounds in the balanced equation. If not, use the ratio from the equation to find of which compound there is too little: that is the limitting reagent, and you must base the rest of your unit analysis on the data for that compound. This may be a little confusing at first...have you done limiting reagent questions in class?

 

[parwana]

whats the equation i will use to find the limiting reagent?

So there are .07 mol of Pb(NO3)2
and .057 of LiBr

Now what

 

[slepsta]

Now you know that LiBr is the limiting reagent. To find theoretical yield balance the chemical equation: Pb(NO3)2 + 2LiBr ---> PbBr2 (s). This tells us that two moles of lithium bromide reacts to give lead (II) bromide. Thus, moles of PbBr2 = .5 moles of LiBr. So the theoretical yield of PbBr2 would be .5 X .0571 moles = .2855 moles. Actual yeild is moles of PbBr2 divided by the molecular weight of PbBr2. Then percent yeild is theoretical moles divided by actual moles times 100%.

 

[Sirus]

Right, although you should really be balancing the whole equation; it doesn't make a difference here, but sometimes it does.
Pb(NO_{3})_{2}+2LiBr\longrightarrow{PbBr_{2}+2LiNO_{3}}

 

[Sirus]

Actually, slepsta I believe you are not entirely correct. Actual yield (mass) is moles of PbBr_{2} multiplied by its molecular weight, not divided, since molecular weight is in \frac{g}{mol}. Also, percent yield is actual yield divided by theoretical yield times 100, as I indicated in post #2.

 

[slepsta]

You are correct Sirus. I meant to say Actual yield (in moles) is mass of ppt PbBr2 divided by its molecular weight.

 

[parwana]

iam getting big numbers, can someone do this in steps and find the answer?

 

[chem_tr]

For the reaction Pb(NO3)2 (aq) + LiBr PbBr2 precipitates out.
Calculate the percent yield (%) if 8.992 g of PbBr2 precipitates when 1.350 L of 0.0550 M Pb(NO3)2 (aq) and 1.400 L of 0.0408 M LiBr (aq) are mixed.

Okay, here goes my attempt:

• First find how many moles are there in 8.992 g of \displaystyle PbBr_2, by using Pb=207.2, and Br=79.9 g/mol.
• Multiply the volumes and molarities of the solutions you mixed to learn their mole amounts.
• If the reaction had proceeded in 100%, you'd find the mole amount identical to \displaystyle Pb(NO_3)_2. So treat this value as 100% yield.
• Divide the first value you found for lead bromide to the value you found for 100% reaction, and multiply the result with 100 to learn the actual percentage yield.

I did those and found 33%. Remember, if your teacher asks you how to improve the yield, your answer must be this; you need to do something about nitric acid formed; so add some dilute solution of sodium acetate or better, sodium carbonate solution to react with it, and according to La Chatelier's principle, the equilbrium has to shift to the products' side.