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Topic: Marathon Problem  (Read 24743 times)

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Offline sapphiregirl

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Re: Marathon Problem
« Reply #15 on: September 03, 2009, 06:54:38 PM »
is it CaC2? Because that would be a total of 64 amu. C2 has 24 amu and 24/64 is 37.5%.

How would you name CaC2? Calcium what?

Offline sapphiregirl

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Re: Marathon Problem
« Reply #16 on: September 04, 2009, 02:08:25 AM »
Is the answer 18.2 grams of C2H2?

This is the final equation I got:

CaC2 + 2H2O -> C2H2 + Ca(OH)2

is that correct?

Offline Borek

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Re: Marathon Problem
« Reply #17 on: September 04, 2009, 02:34:14 AM »
How would you name CaC2? Calcium what?

Carbide or acetylide.

Is the answer 18.2 grams of C2H2?

No, correct answer was already listed earlier in the thread.
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Offline DrCMS

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Re: Marathon Problem
« Reply #18 on: September 04, 2009, 07:19:49 AM »
As Borek said CaC2 is calcium carbide and there are 16.6g acetylene. 
Go look up the derivation of the word limelight and "being in the limelight"

Offline sapphiregirl

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Re: Marathon Problem
« Reply #19 on: September 04, 2009, 11:56:06 AM »
ok, well if it's 16.6 then it can't be CaC2.

But the formula is MC2, yes?

Offline sapphiregirl

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Re: Marathon Problem
« Reply #20 on: September 04, 2009, 12:00:45 PM »
1826, popular name for Drummond light, a brilliant light created by the incandescence of lime (1), adopted for lighthouses and later for the Victorian stage, where it illuminated the principal actors, hence the figurative sense of "on stage, at the center of attention" (1877).

how does that help?

Offline sapphiregirl

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Re: Marathon Problem
« Reply #21 on: September 04, 2009, 12:02:56 PM »
CaO is burnt lime but that only gives you 23% carbon.

Offline Arctic-Nation

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Re: Marathon Problem
« Reply #22 on: September 04, 2009, 12:12:06 PM »
ok, well if it's 16.6 then it can't be CaC2.

But the formula is MC2, yes?
Then your calculations are wrong. All correct answers have already been posted.

Offline Borek

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Re: Marathon Problem
« Reply #23 on: September 04, 2009, 12:27:36 PM »
ok, well if it's 16.6 then it can't be CaC2.

23 grams of B.
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Offline sapphiregirl

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Re: Marathon Problem
« Reply #24 on: September 04, 2009, 12:28:49 PM »
i'm not looking for just the answer. i need to show all my work of how i got to the answer.

so, is it or is it not CaC2? because, i don't know what else substance A could be.

also, how does the forty million trillion formula units have a mass of 4.26 milligrams help at all?

Offline Borek

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Re: Marathon Problem
« Reply #25 on: September 04, 2009, 12:48:43 PM »
also, how does the forty million trillion formula units have a mass of 4.26 milligrams help at all?

Allows calculation of molar mass.

Yers, it can be CaC2. 45 grams were mixed with 23 grams of water. THINK.
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Offline sapphiregirl

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Re: Marathon Problem
« Reply #26 on: September 04, 2009, 12:55:51 PM »
is it a limited reaction problem?

Offline sapphiregirl

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Re: Marathon Problem
« Reply #27 on: September 04, 2009, 01:02:08 PM »
ok, I reworked it and this is what i got.

C is 3/8 of the mass and C is covalently bonded so it's MC2 which means the metal is CaC2 (molar mass of 64)

Here is the final stoich chart I made

BE    CaC2 + 2H2O -> C2H2 + Ca(OH)2
MR   1          2           1          1
MM   64        18          26        74
G     45        23         
Mol  .70        1.3         

so, there's an excess of the CaC2, it's only .65 moles, .05 moles excess (.65 x 2 = 1.3)
so, for the C2H2 we have: .65 moles = g/26 so, g=16.9

but earlier it said the answer was 16.6, so now i'm confused again.

where am i going wrong in my calculations?

Offline Arctic-Nation

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Re: Marathon Problem
« Reply #28 on: September 04, 2009, 01:13:36 PM »
If you go from 1.3 to 1.28 (one more significant number) the result becomes 16.6.

Offline sapphiregirl

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Re: Marathon Problem
« Reply #29 on: September 04, 2009, 01:26:08 PM »
i figured it out. thanks for all your *delete me* i just rounded earlier on so i rounded 1.278 to 1.3 which when you divide by 2 gives you .65 and when you multiply that by 26 you get 16.9 instead of 16.64. thanks again!

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