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Topic: Molarity and grams  (Read 7754 times)

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Offline Geddoe

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Molarity and grams
« on: October 11, 2009, 12:14:57 AM »
Citric acid, H3C6H5O7, reacts with potassium hydroxide to produce potassium citrate and water. How
many grams of citric acid are required for complete reaction with 20.0 mL of 0.200 M potassium hydroxide?
The molar mass of citric acid is 192.125 g/mol. (Hint: start by writing the balanced equation for the
reaction.)


Here is my work:

 H3C6H5O7 + 3 KOH => C6H5K3O7 + 3 H2O

M = n / v

n = M * v
= (0.200 M KOH * 20.0mL) * (1 L / 1000mL) = 0.004 mole KOH

0.004 mole KOH * (1 mole H3C6H5O7  / 3 mole KOH) * (192.125 g/mol / 1 mole H3C6H5O7) = 0.256166667g H3C6H5O7


Is this correct?
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Offline cliverlong

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Re: Molarity and grams
« Reply #1 on: October 11, 2009, 02:20:32 AM »
You have made the same mistake in your other, highly similar, post

moles is not equal to mol / mL * L / mL

Clive

Offline Geddoe

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Re: Molarity and grams
« Reply #2 on: October 11, 2009, 06:50:28 PM »
How can I correct this mistake? In the other topic I believe I did it correctly...  :-\
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Offline Borek

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Re: Molarity and grams
« Reply #3 on: October 12, 2009, 03:46:33 AM »
You are OK, Clive is wrong. I suppose he misread your post and missed fact that M is mol/L.
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Offline cliverlong

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Re: Molarity and grams
« Reply #4 on: October 12, 2009, 05:36:57 AM »
I made the same mistake as in my other post on the related question from Geddoe

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