[Laker12]

A 15 mL sample of .20 M MgCl2 is added to 45 mL of .40 AlCl3. What is the molarity of the Cl- ions in the final solution?

Here's what I was thinking: Seeing as you have mL, you can turn that into liters and then since you have the Molarity you can find the moles of each substance. The problem that I'm facing is that I'm not sure on how to use the given information to find TOTAL molarity of the Cl-. At first I thought that maybe I should find the moles of each and then go from moles of that to moles of Cl-, but then I realized I didn't have the equation. Any help would be greatly appreciated.

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How many electrons are present in 2.0 x 10 ^ -3 moles of Oxygen with a mass number of 18, an atomic number of 8, and a 2- charge?

Ok, now I understand that that the oxygen has 2 extra electrons in this case since the mass number should 16 and there should be 8 protons and 8 electrons. The part that gets me, is how do those 2 extra electrons come into play when you're trying to figure out the total electrons in what they asked? Now the way I see it, since they gave you the moles of oxygen you could use avagadros to find a lot of stuff but I don't think electrons is one of them.

Thanks.

[Borek]

First of all - how many chloride anions per molecule of MgCl₂? Of AlCl₃? How many chloride anions per mole of MgCl₂?

Electrons can be counted using Avogadro numbers. In general Avogadro number is just ovegrown dozen and you can use it for anything - even for cars (although some will state that that's again IUPAC mole definition). For example number of stars in observable universe is in the milimole range

[Laker12]

I've found the Cl anions for MgCl2, so let me run that past you before I do AlCl3 so that I know if I'm doing it right.

1. Turn the given mL into Liters giving you .015 L
2. Multiply Liters by given molarity giving you the grams in the substance - in this case .003. (.015 x .20)
3. Take the given grams of MgCl2 and turn them into moles. (.003/95.21).
4. Take that which I got to be .00003125 and just use it for moles of Cl- since there seems to be one of everything.

I know just by looking at it that I made a mistake in there somewhere.

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Gotcha, so I can use avagadros for this. Question though: Where do those 2 extra electrons come into play when I'm figuring this out?

When I'm doing the problem would I have to do: 2.0 x 10^-3 moles x (6.022 electrons/1 moles)?

[Borek]

2. Multiply Liters by given molarity giving you the grams in the substance - in this case .003. (.015 x .20)

Think it over.

Gotcha, so I can use avagadros for this. Question though: Where do those 2 extra electrons come into play when I'm figuring this out?

When I'm doing the problem would I have to do: 2.0 x 10^-3 moles x (6.022 electrons/1 moles)?

You are partially right - but you are wrong when it comes to number of electrons per anion. How many electrons are there in NEUTRAL oxygen atom? Hint: what does atomic number tells us about nucleus charge?

[Laker12]

Ah I think I see it. Molarity isn't grams of solute over liters of solution, it's MOLES of solute over liters of solution. Ok, so in that case you have .003 moles of MgCl2 right? And the equation for this is Mg (2+) + Cl (-) --> MgCl2 so therefore the balanced euqation is Mg (2+) + 2Cl (-) ---> MgCl2. Therefore you do .003 Moles MgCl2 x (2mol Cl-/1 mol MgCl2). Did I get that step right?

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In a neutral atom there are 8 electrons right? Same as number of protons.

[Borek]

Ok, so in that case you have .003 moles of MgCl2 right?

3 mmoles, right.

And the equation for this is Mg (2+) + Cl (-) --> MgCl2 so therefore the balanced euqation is Mg (2+) + 2Cl (-) ---> MgCl2.

You are going in circles, but at least you are not lost

Therefore you do .003 Moles MgCl2 x (2mol Cl-/1 mol MgCl2). Did I get that step right?

If it leads you to 6 mmoles of Cl^- - yes

In a neutral atom there are 8 electrons right? Same as number of protons.

OK, now what about O^2-?

[Laker12]

Good, so now back to the original question. I have .006 moles of Cl- from MgCl2.

Going by what we did above, I then have .018 moles AlCl3 from the given information. So repeating the steps from the previous one, it'll be .018 moles AlCl3 x (3 moles Cl-/1 mol AlCl3) which gives you a grand total of .054 moles.

So from MgCl2 you have .006 moles of Cl- and from AlCl3 you have .054 moles of Cl-. Seeing as you're trying to find the total molarity, do you then add up the mL given in the problem to get a total of .06 L and then add the total moles you have to get .06 moles and then divide the two since it's moles/L=Molarity to get a total of 1.0 M?

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So if there are 8 electrons in a neutral Oxygen then there are 10 electrons in O(2-).

[Borek]

total of 1.0 M?

Bingo

So if there are 8 electrons in a neutral Oxygen then there are 10 electrons in O(2-).

Bingo again.

[Laker12]

Wonderful, got it. You're the man...again.

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Ok, so 10 electrons in O(2-). How does that affect the process to find the electrons present though?

If you you're setting it up I know you're gonna do 2.0 x 10^-3 moles x a certain amount of electrons over moles, but how do you know how many?

[Borek]

Ok, so 10 electrons in O(2-). How does that affect the process to find the electrons present though?

How many electrons in dozen anions? Now - remembering that mole is just an Avogadro number of atoms/ions/molecules/whatever - how many electrons in a mole of anions?

[Laker12]

I'm getting that 'stupid feeling' again.

There's 6.022 x 10^23 electrons in a mole? I still don't get how that helps set up the problem though.

[Borek]

1 anion of oxygen contains 10 electrons, right?

dozen anions contains 120 electrons, right?

1 mole - 6.02*10²³ anions - contain ...

And don't forget that's 1 mole, and you have fraction of the mole only...

[Laker12]

Alright I think I got it. So:

2.0 x 10^-3 moles of O (2-) x (6.022 x 10^23 anions /1 mole) x (10 electrons/ 1 anion) = 120.44 x 10 ^(-20) or 1.2 x 10^(-22)

right?

[Borek]

Almost OK. Check your math - you should have billions of billions of electrons, not a very tiny fraction of one

[Laker12]

Oops. no '-' in the exponent right?