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Author Topic: How are molar calculations performed?  (Read 2001 times)

Offline dg2008

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How are molar calculations performed?
« on: 20/01/2009 21:36:34 »
2XeF4+ 2H2 → Xe + 4HF

Trying to work out the moles of Xe & HF when 50 cm3 of Xe are produced.
So far I have:

24.5dm3= 24.5x103cm3
24.5x103 cm3 = 1.0 mole of Xe
50cm3 = (1.0/24.5x103 cm3) x 50.0 = 2x103 = 0.002 mole
1 mole of Xe = 4 mole of HF

So is the volume of HF just 4x that of Xe i,e 200cm3?
« Last Edit: 20/01/2009 22:34:50 by chris »


 

Offline lancenti

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How are molar calculations performed?
« Reply #1 on: 21/01/2009 03:39:24 »
First off, the equation's not balanced. You ALWAYS ALWAYS have to check the equation.

XeF4 + 2H2 → Xe + 4HF

Then if we assume that everything in the equation is an ideal gas, then we can simply conclude that for every 1 cm3 of Xe produced, there will be 4 cm3 of HF produced, but that doesn't answer your question of the number of moles. However, it does tell you that you have 200 cm3 of HF

I would do the calculations this way:

Volume of Xe Produced = 50.0 cm3
Amount of Xe Produced = (50.0/1000)/24.5 = 0.002041 mol
Amount of HF Produced = 4*(0.002041) = 0.008163 mol
 

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How are molar calculations performed?
« Reply #1 on: 21/01/2009 03:39:24 »

 

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