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Topic: I need some help with Molarity and solutions cna some one help ASAP  (Read 14095 times)

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Offline CStrike18

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Re: I need some help with Molarity and solutions cna some one help ASAP
« Reply #15 on: March 30, 2008, 06:34:47 PM »
all right now i am really confused... first how i do i find the amount of mols that react with the .02 mols of FeCl3... then you lost me with the molar concentration
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Offline Borek

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Re: I need some help with Molarity and solutions cna some one help ASAP
« Reply #16 on: March 30, 2008, 06:54:52 PM »
Quote from: CStrike18 on March 30, 2008, 06:34:47 PM
all right now i am really confused... first how i do i find the amount of mols that react with the .02 mols of FeCl3...

Simple stoichiometry... You already have balanced reaction equation. Take a look at these pages:

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations
http://www.chembuddy.com/?left=balancing-stoichiometry&right=ratio-proportions

Your balanced reaction equation tells you that you need 6 moles of HCl to prepare 2 moles of FeCl3. If so, how many moles are required to make 0.02 moles of FeCl3?

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then you lost me with the molar concentration

Write down definition of molar concentration and solve it for volume.
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Offline Arkcon

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Re: I need some help with Molarity and solutions cna some one help ASAP
« Reply #17 on: March 30, 2008, 06:56:16 PM »
The definition of Molarity is what?
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Offline CStrike18

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Re: I need some help with Molarity and solutions cna some one help ASAP
« Reply #18 on: March 30, 2008, 07:30:59 PM »
you need .06 moles of HCl

 so then i convert .06 moles to volume
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Offline CStrike18

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Re: I need some help with Molarity and solutions cna some one help ASAP
« Reply #19 on: March 30, 2008, 07:54:22 PM »
like this:
                      36.458g HCl            1 L         
.06 mols HCl X -------------- X  -------------  =  1.99 L
                       1 mol                 1.098 kg/L
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Offline enahs

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Re: I need some help with Molarity and solutions cna some one help ASAP
« Reply #20 on: March 30, 2008, 07:57:55 PM »
Nope.

0.06 mols HCl =   6.00 mols HCl * X L
                            1 L
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Offline CStrike18

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Re: I need some help with Molarity and solutions cna some one help ASAP
« Reply #21 on: March 30, 2008, 08:05:41 PM »
sorry but i am only in 10 grade chem. i really have no idea what those symbols stand for
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Offline Arkcon

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Re: I need some help with Molarity and solutions cna some one help ASAP
« Reply #22 on: March 30, 2008, 08:11:53 PM »
enahs:has written out for you, the calculation needed, to use the definition of molarity, to find the unknown, from known values given to you.  He used the strike though font, for the canceled units.
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Offline enahs

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Re: I need some help with Molarity and solutions cna some one help ASAP
« Reply #23 on: March 30, 2008, 08:13:19 PM »
Heh, yeah. They do look like some kind of symbol. It is just as Arkcon said, strike-through fonts, and in this case the units of Liters cancels.

Always follow then units.

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Offline CStrike18

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Re: I need some help with Molarity and solutions cna some one help ASAP
« Reply #24 on: March 30, 2008, 08:16:19 PM »
oh i see thank you for clarifying that for me
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Offline CStrike18

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Re: I need some help with Molarity and solutions cna some one help ASAP
« Reply #25 on: March 30, 2008, 08:22:43 PM »
Quote from: enahs on March 30, 2008, 07:57:55 PM
Nope.

0.06 mols HCl =   6.00 mols HCl * X L
                            1 L

i got 36 mL for my answer
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Offline Borek

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Re: I need some help with Molarity and solutions cna some one help ASAP
« Reply #26 on: March 31, 2008, 02:55:30 AM »
Quote from: CStrike18 on March 30, 2008, 08:22:43 PM
Quote from: enahs on March 30, 2008, 07:57:55 PM
Nope.

0.06 mols HCl =   6.00 mols HCl * X L
                            1 L

i got 36 mL for my answer

Nope. Solve equation above for X.
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Offline CStrike18

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Re: I need some help with Molarity and solutions cna some one help ASAP
« Reply #27 on: March 31, 2008, 08:56:16 PM »
i figured it out last night after signing off... it was 10mL... thanks for the help every one
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