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Topic: Lithium - Moles and RAM  (Read 4682 times)

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mattgad2006

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Lithium - Moles and RAM
« on: March 04, 2006, 12:05:57 PM »
Hi All,

I did an experiment to determine the RAM of lithium using 0.12g of lithium, by measuring the hydrogen produced when it reacted with water.

222cm3 of H2 was given off.

Assuming 1 mole of gas occupies 24000cm3 at room temperature and pressure, I have worked out that the number of moles of hydrogen collected was 0.00925 moles and that the moles of Li that reacted is 0.01731913043. Are these calculations correct?

How do I use this information to calculate the relative atomic mass of lithium?

The equation for the reaction is: 2Li(s) + 2H2O(l) --> 2LiOH(aq) + H2(g)

Thanks.
« Last Edit: March 04, 2006, 12:15:32 PM by mattgad2006 »

Offline Alberto_Kravina

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Re:Lithium - Moles and RAM
« Reply #1 on: March 04, 2006, 12:27:38 PM »
I think you are correct!

1 mole ----> 24000 cm3
x moles -----> 222 cm3

X=(1*222)/(24000)
X = 0.00925 moles
« Last Edit: March 04, 2006, 12:28:31 PM by Alberto_Kravina »

mattgad2006

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Re:Lithium - Moles and RAM
« Reply #2 on: March 04, 2006, 12:46:48 PM »
Thanks, am I also correct for the moles of Li?

Offline xiankai

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Re:Lithium - Moles and RAM
« Reply #3 on: March 05, 2006, 06:47:40 AM »
how did u arrive at that number? all u got to do is multiply moles of hydrogen by 2 according to the mole ratio, why are there so many digits?
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