Chem question

[grayfalcon89]

I have two questions on chemistry. We recently did a lab on MOLE RATIO AND CHEMICAL REACTION. What we did was observed the precipitate that was created from combining [math]Pb(NO_3)_2[/math] and [math]Na_2 CO_3[/math]. The balanced equation for this chemical reaction was:

 

[math]Pb(NO_3)_2 + Na_2 CO_3 \rightarrow PbCO_3 + 2NaNO_3[/math]

 

Okay.. So, using limiting reactant and other things, I was able to predict the number of moles of [math]PbCO_3[/math] that was resulted. But I can't find the answer to two questions that followed it (I have approaches but I don't know if this is right..). We used [math]0.50 M[/math] for molarity (the concentration of solution).

 

1. Describe how you would verify your predictions.

 

My answer: I will go back to my lab station and find the volume of the precipitate in ml. This is easy since the sum of the volumes of [math]PbCO_3[/math] and [math]NaNO_3[/math] is 8 throughout the lab. Now, converting 8.00 ml into liters, we get 0.008 l. Since the product of molarity and liters equal to the number of moles, we can find the number of moles by multiplying 0.008 liters and 0.50 M to get 0.004 moles.

 

I thought this worked out but my teacher looked at me puzzled on this. Also, my friend must've did something else because he told me that he used [math]M_1V_1 = M_2V_2[/math] or something. Any idea on this one?

 

2. Though you measure precipitate heights in this experiment, what property of the precipitates were you measuring indirectly?

 

My answer: I'm guessing it's volume because although we did write the height in mm by ruler, we used the volume of each [math]Pb(NO_3)_2[/math] and [math]Na_2CO_3[/math] in beginning to form precipitate so I'm thinking that the answer is VOLUME..

 

Thanks for all your help! And go Latex too!

[woelen]

Indeed, you have to do what your friend suggested. The TOTAL volume is 0.008 liter, but the volume of Pb(NO3)2 solution and the volume of Na2CO3 solution is lower (e.g. 2 ml and 6 ml). So, you compute how many moles of Pb(NO3)2 you have from the original volume of that solution and you compute how many moles of Na2CO3 you have from the original volume of solution of Na2CO3.

 

Now, when you have the number of moles for both compounds, you can determine, which is the limiting reagent and then you can compute the amount of PbCO3 formed in the reaction.

 

The second question is not clear to me. What do you mean with "precipitate height"? Is that the total height of the liquid? The only way to measure the precipitate correctly is to filter it out of the solution, rinse it carefully and dry it carefully, without loosing any precipitate and weighing the amount of precipitate. If with measuring the precipitate height you mean measuring the height of the spongy or jelly like structure in the liquid, then I think that you are doing a useless thing. The density of the precipitate may vary wildly.

[grayfalcon89]

Okay.

 

This is so peculiar that I have to ask about it...

 

From [math]M_1 V_1 = M_2 V_2[/math], I assume the molarity for both [math]M_1,M_2 = 0.50[/math]. So, letting x = V_1, the problem trivializes because this gives x = 0.008 - x which gives x = V_1 = V_2 = 0.004 liters. To find the numbe of molarity, we multiply this by 0.5 which gives 0.002 moles. Now, here is the peculiar part. The coefficient is same and the number of moles is same.. So, there is no limiting reactant!!!

 

Anyway, if I set one of them to be the "known" and [math]Pb(CO_3)[/math] be the unknown, I can set up the ratio:

 

[math]1/0.002 = 1/y[/math]

 

where y is the number of moles for [math]Pb(CO_3)[/math]. So, the predicted number of moles is 0.002.

 

But this is weird because if you solved for [math]NaNO_3[/math], what happens it that you get 1/0.002 = 2/z and this tells z = the number of moles for [math]NaNO_3[/math] = 0.004. But then this means the products have 0.002 + 0.004 = 0.006 moles and by the Law of Conservation of Matter, this is a contradiction. What did I do wrong?

[jdurg]

The law of conservation of matter only deals with the conservation of mass, not the number of moles. Look at the equation for the formation of ammonia via the Haber process. 3H2 + N2 => 2NH3. There you have 4 moles becoming two moles. The mass is the same but the number of moles is different. This is perfectly acceptable.

[grayfalcon89]

The law of conservation of matter only deals with the conservation of mass, not the number of moles. Look at the equation for the formation of ammonia via the Haber process. 3H2 + N2 => 2NH3. There you have 4 moles becoming two moles. The mass is the same but the number of moles is different. This is perfectly acceptable.

 

 

Oh okay. Oh by the way, was my work right? (The part about finding the moles of two reactants being 0.002 mol; I wasn't sure about this because I didn't get limiting reactant. :-O)

[woelen]

Yes, that is correct. In this situation, the reactants match precisely, so nothing remains unreacted.

 

You will get a limiting reagent if you change the ratio, e.g. 0.003 l of Pb(NO3)2 and 0.005 l of Na2CO3. Just do the math and determine, which is the limiting reagent, how many moles of the reaction product you get and how many moles of the other reagent remains.

 

As another exercise for you to see if you really understand. Replace the solution of 0.5M Na2CO3 by a solution of 0.5M NaI.

 

With NaI you also get a similar reaction:

 

Pb(NO3)2 + 2NaI --> PbI2 + 2NaNO3

 

Here PbI2 precipitates from solution (a yellow precipitate). Again, assume total volume is 0.008 l. Determine how much solution of Pb(NO3)2 and how much solution of NaI is needed to have precisely no limiting reagent.

[grayfalcon89]

YESSS!!

 

Thank you!