[Newt]

This is a cool forum! Glad to be a a part of it.

OK, here's my questions. I was sick yesterday and friday when my teacher went over how to calculate this.

First off, we are supposed to write out the balanced equation for the reaction of lithium carbonate and hydrochloric acid...?

Here is what I got:

Li2CO3 + 2HCl  --->  2LiCL + CO2 + H20

I think this is correct??

Now for this reaction, we are supposed to answer the following:

1. Identify (and name) the excess reagent and tell what happens to the excess reagent...

I am guessing here, but am thinking the excess reagent would be Li2CO3 (lithium carbonate).

But I am not really sure how to calculate and determine the excess and limiting reagent...and then am not sure how to answe the question of "what happens to the excess reagent"?

Can anyone figure that one out?

The other two questions are as follows:

2. Calculate the mass of NaCL that is produced if 1.5000 grams of Na2CO3 reacts with excess HCl.

I wrote the reaction as follows (hope it is correct):

Na2CO3 + 2HCl  ---> Co2 + H2O + 2NaCl

Since I missed class, I am not really sure how to calculate the mass of NaCL produced in the above reaction.

3. Calculate the mass of CaCO3 that would be required to produce 1.0000 grams of CaCl2.

Here is how I set up the reaction:

CaCO3 + 2HCL  --->  CaCl2 + CO2 + H2O

Again, not sure how to set up the problem to find the mass of the reactant CaCO3.

4. If 4.000 Moles of HCL was available to use, determine the volume of 4.000 Moles of HCl solution required to completely react with 2.000 grams of Na2CO3.

The answer I get is approximately 9.435 mL of 4.000 HCl solution.

here's how I got the answer:

I took the given amount of 2.000 g of Na2CO3 and multiplied that by (1 Mol of Na2CO3 over molar mass of Na2CO3) to get approximately .01887 moles

than I took .01887 moles of Na2CO3 and multiplied that number by (2 Mol HCL over 1 Mol Na2CO3...or 2/1) to get approximately .03774 Mol of HCl.

I then took ,03774 Mol HCl and multiplied that by 1.00 Liter of solution over 4.000 Mol HCl to get .009435 HCl

I then multiplied .009435 by 1000 ml (because 1 Liter=1000 mL and I want the answer in Ml, not Liters) to give me approximately 9.435 mL of HCl solution

Is that correct answer and the right way to do it?

Can anyone help me on questions 1, 2 and 3, and let me know if my answer and method for question 4 is correct?

Thanks!!

Newt


                 

[Alberto_Kravina]

Li2CO3 + 2HCl  --->  2LiCL + CO2 + H20
I think this is correct?

Correct!

Na2CO3 + 2HCl  ---> Co2 + H2O + 2NaCl

Correct!
It's quite simple to calculate how much NaCl is formed:

First of all, calculate the number of moles of Sodium carbonate with the formula n=m/M

n(Na₂CO₃) : n (NaCl) = 1 : 2
=>2*n(Na₂CO₃)=n(NaCl)

so now you know how many moles of NaCl are formed. After this you can calculate the mass with the formula n=m/M (in this case you must calculate the mass "m")

-show us your results- my result is 1,654166429 g NaCl

[Newt]

Correct!Correct!
It's quite simple to calculate how much NaCl is formed:

First of all, calculate the number of moles of Sodium carbonate with the formula n=m/M

n(Na₂CO₃) : n (NaCl) = 1 : 2
=>2*n(Na₂CO₃)=n(NaCl)

so now you know how many moles of NaCl are formed. After this you can calculate the mass with the formula n=m/M (in this case you must calculate the mass "m")

-show us your results- my result is 1,654166429 g NaCl

yay!! Yes, thank you!! I believe I got the same answer!!

I found the molar mass of 1 M of Na2CO3 to be 105.9884

I found n of Na2CO3 to be 1.5000/105.9884= .01452 M of Na2CO3

Thus, since the equation is in a is a 1:2 M ratio (Na2CO3 ---> 2NaCl or 1:2 ), I then multiplied 0.01452 x 2 = 0.028304

To find the mass of NaCL:

0.02834 = 1 M NaCL/ 58.44277 g

= 0.02834 x 58.44277 g
= 1.65416416208 g NaCL
= approximately 1.6542 g of NaCL

Is this the right method to calculat the amount of g of NaCL produced??

Thanks for your *delete me*! Any ideas about questions 1, 3 and 4 in my original post?  As for question 3 (a question asking me to determine the amount of reactant required to produce a certain amount of product), do I use the same method to find the mass of reactants as I did to find the mass of the product??

Please let me know. Thanks again fr your *delete me*!

[Albert]

As for question 3 (a question asking me to determine the amount of reactant required to produce a certain amount of product), do I use the same method to find the mass of reactants as I did to find the mass of the product??

Yes, go on: you're on the right track.

[Newt]

Yes, go on: you're on the right track.

Thanks, I'm trying

Ok, so for question 3 in my original post:

3. Calculate the mass of CaCO3 that is required to produce 1.0000 g of CaCl2?

So, my balanced equation was as follows:

CaCO3 + 2HCl ---> CaCL2 + CO2 + H2O

I determine this to be in a 1:1 ratio (1 M CaCO3---> 1 M CaCl2)

So then, I took the given amountof 1.000 g of CaCl2 and determined the number of Moles (M) of CaCl2 by dividing the 1.5000 g by the molar mass of 1 M CaCl3 (or 75.531g):

1.5000 g CaCl2 / 110.984 g CaCl2
= approximately 0.0090103 M CaCl2

Since it is a 1:1 M ratio, I determined the mass of of CaCO3 to be as follows:

0.0090103  = 1 M CaCO3 / 100.0869 g CaCO3

= 0.013240 x 100.0869
= 1.325150556 g CaCO3
= approximately 0.90181 g CaCO3

So, approximately 0.90181 g of CaCO3 is required to produce 1.0000 g of CaCl2.

Is this what you get??

Any thoughts on question 1 and 4 in my original post?? Your help is helping!!
Thanks!  

[Albert]

Any thoughts on question 1?

Check this out: http://www.science.uwaterloo.ca/~cchieh/cact/c120/limitn.html

If 4.000 Moles of HCL was available to use

Do you mean a 4M (moles/Liter) solution? Because, if that's how things are, you found the right answer.

[Newt]

Check this out: http://www.science.uwaterloo.ca/~cchieh/cact/c120/limitn.htmlDo you mean a 4M (moles/Liter) solution? Because, if that's how things are, you found the right answer.

Yes, 4 M (moles/Liter) solution. Great!!

Thanks for the link on reagents; checking it out now.

[Newt]

Check this out: http://www.science.uwaterloo.ca/~cchieh/cact/c120/limitn.htmlDo you mean a 4M (moles/Liter) solution? Because, if that's how things are, you found the right answer.

Thanks for the link. After looking at it, I am still somewhat confused.

I''ve reread my question/homework problem, which is as follows:

In the following reaction:

Li2CO3 + 2HCl --->  2 LiCl + CO2 + H2)

1a) What is the excess reagent in the experiment?  1b) What happens to the xcess reagent at the end of the experiment?

One problem, as I see it, is that I am NOT given any information concerning how many grams of lithium carbonate and hydrochloric acid are in the experiment, other than the balanced equation above.

Looking at the balanced equation, it appears that the reaction requires 1 M of Li2CO3 with 2 M of HCl...
or a 1:2 ratio.

So, does that mean then that HCL is the excess reagent, because it requires more Moles of HCL in the reaction?

BUT, in comparing the masses, I calculate that

1 M of lithium chloride is approximately 73.8909 g (this is more than the molar mass of 2 M HCl)

AND

2 M HCl is approximately 72.92188 g (this is less than the molar mass of 1 M Li2CO3)

So, now I am more confused lol I think this would be a good Double Jeopardy question

Any other hints on this one? :d