[mac227]

We took 3mL of a .5 M solution of CaCl2 and a .5 M solution K2CO3 and mixed them.  A white precipitate--CaCO2--formed.  

CaCl2 + K2CO3   CaCO2 (s) + 2K⁺ + 2Cl^-

I need to determine the mass of starting material and product.

Mass of reactants:

.5 mol/L (L/1000mL)  (3mL/1)  (138.18 g K2CO3/1 mol K2CO3) = .20727 g K2CO3
.5 mol/L (1L/1000mL) (3mL/1)  ( 110.9g CaCl2 / 1mol) = 0.16635 g/mL CaCl2

mass ratio is 40:60      empirical formula is 2:3     i believe..


the moles of reactans would be 1.5 * 10^-3 for both K2CO3 and CaCl2

Now i need to find the mass and moles of product CaCO2.  


2K and 2Cl are spectator ions so i do not need to include their mass and moles correct??

How do I go about calculating mass and moles of the product CaCO2??

[opti384]

The balanced equation tells us that when 1 mol of CaCl2 and 1 mol of K2CO3 react, 1mol of CaCO2 and 2 mols of KCl are produced.

[mac227]

The balanced equation tells us that when 1 mol of CaCl2 and 1 mol of K2CO3 react, 1mol of CaCO2 and 2 mols of KCl are produced.

So K and Cl are not spectator ions??  I thought they dissociate in water??

So is my equilibrium incorrect? Should I be calculating the molar mass of CaCO2 and KCl as my products?

[opti384]

Indeed K and Cl are spectator ions. The process which you used to find the determine the mass of the starting materials is also correct.

What I am saying here is that if each 1.5 * 10-3 mols of CaCl2 and K2CO3 react, there will be 1.5 * 10-3 mols of CaCO2 and 3.0 * 10-3 mols of KCl.

[mac227]

Indeed K and Cl are spectator ions. The process which you used to find the determine the mass of the starting materials is also correct.

What I am saying here is that if each 1.5 * 10-3 mols of CaCl2 and K2CO3 react, there will be 1.5 * 10-3 mols of CaCO2 and 3.0 * 10-3 mols of KCl.

So even though they are spectator ions I cannot ignore them? 

I am confused why KCl is 3.0 * 10 -3     is it just adding 1.5 * 10-3 K   +    1.5*10-3 Cl??

And the mass of product would be 1 mol CaCO2 (84 g / 1 mol) = 84 g CaCO2??

And I would have to also include the mass of the spectator ions individuals or as KCl??

I figure since they are spectator ions i would ignore them, but if not then I would justs start out with 1.5*10 -3 mol K and use that to get the mass in grams?? (same goes for Cl)

[opti384]

I think in this case you can just ignore the spectator ions.

So the overall reaction will be

CaCl2 (aq) + K2CO3 (aq)  CaCO2 (s) + 2KCl (aq)

so 2KCl (aq) and 2K⁺ + 2Cl^- basically mean the samething.

And the mass of product would be 1 mol CaCO2 (84 g / 1 mol) = 84 g CaCO2??

as you look up at the reaction, CaCl2 and K2CO3 react and form CaCO2 on a 1:1:1 ratio. So since you have

1.5 * 10-3 mols of CaCl2 and K2CO3 each, there will be 1.5 * 10-3 mols of CaCO2.

[mac227]

I think in this case you can just ignore the spectator ions.

So the overall reaction will be

CaCl2 (aq) + K2CO3 (aq)  CaCO2 (s) + 2KCl (aq)

so 2KCl (aq) and 2K⁺ + 2Cl^- basically mean the samething.

And the mass of product would be 1 mol CaCO2 (84 g / 1 mol) = 84 g CaCO2??

as you look up at the reaction, CaCl2 and K2CO3 react and form CaCO2 on a 1:1:1 ratio. So since you have

1.5 * 10-3 mols of CaCl2 and K2CO3 each, there will be 1.5 * 10-3 mols of CaCO2.

Ok so since we can ignore spectator ions I can calculate the mass of CaCO2 and that will be the mas of the product correct?


1.5 * 10-3 mols of CaCO2  (84/1) = 0.126 g CaCO2


For the sake of it the mass of spectator ions would be 3.0* 10 -3 ( molar mass K2CO3/1) = final mass K2CO3     correct?

or would i need to calculate the mass of them individually ?

[opti384]

1.5 * 10-3 mols of CaCO2  (84/1) = 0.126 g CaCO2

Correct

Now I think you probably need only the mass of CaCO2 for the product, but if you have to also include the

spectator ions in the products, you should add the mass of KCl (aq).

[DrCMS]

Note to both of you calcium carbonate is the product and its formula is CaCO₃

Now try again.

[opti384]

Thanks. It totally slipped out of my mind. How embarrassing.