Need some tips about solving confusing Stoic Problem

[Lori]

Homework Statement

What mass of iron(3) chloride should be added to 450 ml of 0.132 M CaCl2 to give a solution with final chloride concentration of 0.555M

There are a lot of numbers, and I get confused on how to convert it to get mass of iron(3) chloride!

Homework Equations

M = n/L

The Attempt at a Solution

? g iron(3) chloride (162.2 g/mol)
.450 L of 0.132 M CaCl2 gives 0.0594 mols
0.555 M Cl- = 0.555 mols/L

I know the mole to mole ratio we would use is 2:3 but like, i get stuck at trying to start with the right numbers in my dimensional analyst...

my attempt:

0.555 mols/ L Cl- * (0.45 L CaCl2) * (3 mols Cl/2molsCl) *(1mol FeCl3/3molsCl ) = 0.12 (which is wrong).

 

[DoItForYourself]

Firstly, you must find how many mols of Cl^- you must add to the existing solution to achieve chloride concentration of 0,555 M.

Then, you must think how many moles of FeCl₃ are needed to provide these mols.

Finally, using the molar mass you can calculate the respective mass of FeCl₃.

 

[epenguin]

Even before that you'd have to know what is the molarity of chloride in the existing CaCl₂ solution?

 

[DoItForYourself]

Of course,

CaCl₂ disassociates by the reaction:

CaCl₂ $\to$ Ca²⁺ + 2Cl^-.

So, the molarity of chloride is two times the molarity of CaCl₂. Right?

 

[epenguin]

Right enough - so give it a number.

 

[Lori]

Firstly, you must find how many mols of Cl^- you must add to the existing solution to achieve chloride concentration of 0,555 M.

Then, you must think how many moles of FeCl₃ are needed to provide these mols.

Finally, using the molar mass you can calculate the respective mass of FeCl₃.

I figured out that there are 0.1188 mols of Cl- in the solution but how do I know how many moles Cl to achieve .555 M cl?

 

[Lori]

Even before that you'd have to know what is the molarity of chloride in the existing CaCl₂ solution?

Hi. I just started the problem again and I'm still confused. I got the moles of Cl that's in the current solution but I'm not sure how to find how many more moles we need to get the .555 M Cl concentration

 

[Borek]

Typically all you need in such situations is subtraction.

 

[Lori]

Typically all you need in such situations is subtraction.

I know that but isn't. .555 M like .555 mols per liter? How can I get mols so that I can subtract it by .1188

 

[epenguin]

You can think in moles or you can think in molarity which is as you say moles per litre, and usually you will have to go a bit back and forth between them.

There is not just one right way. However I would find just a bit easier since you are given a molarity to start with, and you aim to finish with another molarity, to answer my question with molarity of chloride 0.132 × 2 = 0.264 M. You are asked to achieve final molarity of 0.555 M - so what's the molarity in chloride you need to add - see Borek, and it does that make sense?

 

[Lori]

You can think in moles or you can think in molarity which is as you say moles per litre, and usually you will have to go a bit back and forth between them.

There is not just one right way. However I would find just a bit easier since you are given a molarity to start with, and you aim to finish with another molarity, to answer my question with molarity of chloride 0.132 × 2 = 0.264 M. You are asked to achieve final molarity of 0.555 M - so what's the molarity in chloride you need to add - see Borek, and it does that make sense?

So. I should start with .555 M of chlorine ?

 

[Lori]

I'm am still lost ;(. Here's what I got

 

[epenguin]

Sorry I just can't read that. In future if you must post Photos of homework pages, which is discouraged and certainly not necessary for such a simple problem as this, please use an app like the free DocScanHD to render them readable and post them the right way up.

OK, so for a litre you want there to be 0.555 moles of chloride, you have got 0.264 moles of chloride (see #10) - how many more moles of chloride do you need there to be in that litre? How difficult is that?

 

[I like Serena]

Hey Lori! :)

I have trouble reading what you've got as well, and moreover, I have trouble making sense what everything is, and how they are connected.

To solve a problem like this, it usually helps to write down all given information first, and make a before-after matrix.

Something like:
$$\begin{array}{ll} V=450\text{ mL}\\
c(CaCl_2)=0.132\text{ mol/L}\\
c_{after}(Cl^-)=0.555\text{ mol/L}\\
M(FeCl_3)=162.2\text{ g/mol}\\
\begin{array}{|c|c|c|}
\hline
&Before & After & Delta \\
\hline
c(Cl^-) & ? & 0.555\text{ mol/L} & - \\
n(Cl^-) & ? & ? & ?\\
n(FeCl_3) & ? & - & -\\
\hline
\end{array} \\
m(FeCl_3) =?\end{array}$$
Relevant formulas:
- moles are concentration times volume ($n=c\cdot V$),
- mass is moles times molar mass ($m=n\cdot M$).

Can we find all the question marks?

 

[Lori]

Hey Lori! :)

I have trouble reading what you've got as well, and moreover, I have trouble making sense what everything is, and how they are connected.

To solve a problem like this, it usually helps to write down all given information first, and make a before-after matrix.

Something like:
$$V=450\text{ mL}\\
c(CaCl_2)=0.132\text{ mol/L}\\
c_{after}(Cl^-)=0.555\text{ mol/L}\\
M(FeCl_3)=162.2\text{ g/mol}\\
\begin{array}{|c|c|c|}
\hline
&Before & After & Delta \\
\hline
c(Cl^-) & ? & 0.555\text{ mol/L} & - \\
n(Cl^-) & ? & ? & ?\\
n(FeCl_3) & ? & - & -\\
\hline
\end{array} \\
m(FeCl_3) =?$$
Relevant formulas:
- moles are concentration times volume ($n=c\cdot V$),
- mass is moles times molar mass ($m=n\cdot M$).

Can we find all the question marks?

I'm Sorry, but I expected that there would be confusion cause my calculation is all over the place! I will follow your strategy. I'll let you know what I get!

 

[Lori]

You can think in moles or you can think in molarity which is as you say moles per litre, and usually you will have to go a bit back and forth between them.

There is not just one right way. However I would find just a bit easier since you are given a molarity to start with, and you aim to finish with another molarity, to answer my question with molarity of chloride 0.132 × 2 = 0.264 M. You are asked to achieve final molarity of 0.555 M - so what's the molarity in chloride you need to add - see Borek, and it does that make sense?

Ok, i got that we need 0.291 mols Cl- /literOh i see! We need .291 mols Cl- /liter, but we have 0.450 L of CaCl2, so if we multiply these numbers together , we get the mols we need of Cl-. With 0.13095 mols Cl-, calculate moles of FeCl3 we need by using ration 1 fecl3 to 3 moles cl-. Converting to grams in FeCl3 with molar mass, we get 7.08 grams?

Is it is 7.08 grams of FeCl3? :OI'm pretty sure i spent like 3 hours on this problem, but i realize that i just need to think about and ask the same questions that you guys asked. Sigh!

Thanks for your guys patients >.<

 

[Borek]

7.08 g of FeCl₃ it is.

i just need to think about and ask the same questions that you guys asked

We don't ask them without reason.

Note: in the real lab you would have serious problems finding anhydrous FeCl₃, as this compound is typically sold as hexahydrate (FeCl₃⋅6H₂O). That means you would need almost 12 g of the solid.