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Non Life Sciences => Chemistry => Topic started by: dg2008 on 20/01/2009 21:36:34

Title: How are molar calculations performed?
Post by: dg2008 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?

Title: How are molar calculations performed?
Post by: lancenti 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