[andyman20]

Hi
I'm having trouble with thistopic and I'm very frustrated..

The question is:

9.0mL of 0.10 mol/L NaOH solution is added to 500mL of 1.00 mol/L CdCl2 solution.

                            CdCl2 + 2NaOH ----> 2NaCl + Cd(OH)2

Calculate the mass of Cd(OH)2

I've done everything I could do...

The molecular mass of Cd(OH)2 = 146.42
moles of NaOH = 0.225
moles of CdCl2 = 2.73

how do you find the mass of Cd(OH)2???
please help me...

[arnyk]

Good so far, now you have to find the limiting/excess reagent in order to know which numbers to use.

So if you have 2.73 mol of CdCl2, you would need twice the amount of NaOH (because in the balanced equation it is 1:2).

But that would mean you need 5.46mol of NaOH, which you don't have.  You only have 0.225mol of NaOH.

Therefore, which is the limiting and which is the excess reagent?

Once you know the numbers to use, it is simply a mole ratio to determine the moles of Cd(OH)2.

For example, if you figure out that it takes 2mol of CdCl2 for a complete reaction, then you also know that it will produce 2mol of Cd(OH)2, because they are in a 1:1 ratio in the balanced equation.

Knowing the moles will let you figure out the mass in grams using molar mass.

[andyman20]

thanks for your reply but i sill dont get it...
um.. can u write down the steps of getting the mass and the use of limiting and excess reagent.

Thank you very much

[Mitch]

We do not give answers here strangely. Keep asking us questions until you figure it out yourself, we'll keep pushing you in the right direction.

[arnyk]

I'll do a similar example problem.

Determine the mass of PbS produced when 3mol of Pb3(PO4)2 is reacted with 6mol of K2S?

Pb3(PO4)2 + 3 K2S --> 3 PbS + 2 K3PO4

Ratio of Pb3(PO4)2 to K2S is 1:3, so if you have 3 mol of the Pb3(PO4)2 you would need 9mol of K2S.

But you don't have 9mol of K2S.

So you look that you have 6mol of K2S, meaning you would need 2mol of Pb3(PO4)2 (remember 1:3).  This will work, because you have more than enough Pb3(PO4)2 (you have 3mol when you only need 2) in order to completely react.

Now here's what you use in the reaction:

K2S = 6mol (completely used, therefore it is the limiting reagent)
Pb3(PO4)2 = 2mol (you actually had 3mol so you will have 1mol in excess making this the excess reagent)

Ok so now you know how much of each substance you are using.

Recall the equation:

Pb3(PO4)2 + 3 K2S --> 3 PbS + 2 K3PO4

See that there is a 1:1 ratio between K2S and PbS (3:3 simplifies to 1:1), therefore if you have 6mol of K2S you will produce 6mol of PbS.

Knowing the moles you may now calculate mass.

I find that examples often help the most in understanding the problem thoroughly, however if you did not understand the underlying concept at all during the lesson this will not help.

[hmx9123]

Is this a mistake, K₂I?  Iodine with a -2 charge?  I know iodine has some funky chemistry with the triiodide ion, etc., but are these feasible?  I haven't heard of it before, and it's not in the CRC.  Also, although PbI exists, what redox is going on here?  Shouldn't it be PbI₂?

Wouldn't your overall equation be:

Pb₃(PO₄)₂ + 6 KI --> 3 PbI₂ + 2 K₃PO₄

[arnyk]

Oops.  

Right, it's a halide..1- charge...wasn't quite thinkin' there eh?

Ok then, I guess the easiest way to cover that up would be to change the element.

Wow that's just really embarassing...