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Topic: Naming Compound with Given mass + # of moles in ox/reduction reaction  (Read 6485 times)

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Offline Laker12

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A .242 g sample of potassium is heated in oxygen. The rest is .440 g of a crystalline compound. What is the formula of this compound?

I'm a little confused as to how to use the given information to find the formula. Do I use the molar mass of oxygen to turn into moles? Even then I'm not sure what to do about the rest.

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These three equations describe an oxidation reduction method for determining dissolved oxygen in water. How many moles of S203- (S2O3 with a negative 1 charge) are equivalent to each mole of O2?

1. 2Mn(2+) [aq] + 4OH(-) [aq] ---> 2MnO2 + 2H2O [l]
2. MnO2 + 2I (-) [aq] + 4H (+) [aq] ---> Mn(2+) [aq] + I2 [aq] + 2H2O [l]
3. 2S2O3(2-) [aq] + I2 [aq] ---> S4O6(2-) [aq] + 2I(-) [aq]

To be honest, this one completely puzzles me. I'm at a loss for even which method to use to solve this problem.

Thank you in advance.

Offline Borek

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Re: Naming Compound with Given mass + # of moles in ox/reduction reaction
« Reply #1 on: December 08, 2006, 04:28:41 AM »
A .242 g sample of potassium is heated in oxygen. The rest is .440 g of a crystalline compound. What is the formula of this compound?

Follow the path:

Potas mass. Moles of potas. Mass of oxygen that reacted. Moles of oxygen. Formula.

Quote
These three equations describe an oxidation reduction method for determining dissolved oxygen in water. How many moles of S203- (S2O3 with a negative 1 charge) are equivalent to each mole of O2?

No oxygen in these reactions, no such compound as S2O3-, no wonder you have no idea what to do. Check out first reaction - it is not balanced.
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Offline Laker12

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Re: Naming Compound with Given mass + # of moles in ox/reduction reaction
« Reply #2 on: December 08, 2006, 01:13:57 PM »
I got the first one, thanks.

As far as the second one, that was a mistake on my part. Here's the right info:

Quote
1. 2Mn(2+) [aq] + 4OH(-) [aq] + O2 [g] ---> 2MnO2 + 2H2O [l]
2. MnO2 + 2I (-) [aq] + 4H (+) [aq] ---> Mn(2+) [aq] + I2 [aq] + 2H2O [l]
3. 2S2O3(2-) [aq] + I2 [aq] ---> S4O6(2-) [aq] + 2I(-) [aq]

Sorry about that.


Offline Borek

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Re: Naming Compound with Given mass + # of moles in ox/reduction reaction
« Reply #3 on: December 08, 2006, 03:13:48 PM »
Simple stoichiometry now. 1 oxygen molecule gives 2 molecules of MnO2. 2 molecules of MnO2 give ... molecules of I2. ... molecules of I2 react with ... molecules of thiosulphate.

Just fill the blanks.
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Offline Laker12

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Re: Naming Compound with Given mass + # of moles in ox/reduction reaction
« Reply #4 on: December 08, 2006, 03:34:31 PM »
This is going to sound stupid, but why does 1 oxygen molecule give up 2 molecules of MnO2?

Offline Borek

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Re: Naming Compound with Given mass + # of moles in ox/reduction reaction
« Reply #5 on: December 08, 2006, 03:49:51 PM »
This is going to sound stupid

I am not denying ;)

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but why does 1 oxygen molecule give up 2 molecules of MnO2?

See first reaction equation - how many oxygen molecules on the left? How many MnO2 molecules on the right? That's stoichiometry. See stoichiometric calculations - 2nd paragraph is about reading the balanced reaction equation.
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Offline Laker12

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Re: Naming Compound with Given mass + # of moles in ox/reduction reaction
« Reply #6 on: December 08, 2006, 04:02:48 PM »
I can appreciate honesty. ;)

Now let me see if I understood it correctly from that site. Would 2 molecules of MnO2 produce 2 molecules of I2 since in that second equation there is 1 molecule of MnO2 and 1 molecule of I2?

Offline Borek

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Re: Naming Compound with Given mass + # of moles in ox/reduction reaction
« Reply #7 on: December 08, 2006, 04:08:43 PM »
Would 2 molecules of MnO2 produce 2 molecules of I2 since in that second equation there is 1 molecule of MnO2 and 1 molecule of I2?

Exactly!
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Offline Laker12

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Re: Naming Compound with Given mass + # of moles in ox/reduction reaction
« Reply #8 on: December 08, 2006, 04:13:04 PM »
Beautiful, I got it, you're the man.

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