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Topic: Limiting Reactants  (Read 11686 times)

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  • Guest
Limiting Reactants
« on: October 01, 2004, 07:07:40 PM »
I have searched everywhere in my Chemistry book and so far have been unable to find examples of how to solve the following problem:

GeF3H is formed from GeH4 and GeF4in the combination reaction:

GeH4 + 3GeF4 --> 4GeF3H

If the reaction yield is 92.6%, how many moles of GeF4 are needed to produce 8.00 mol of GeF3H?

I can get the molecular weight of the substances but I am unsure of how to completely solve this problem.  I know the answer is 6.48...it's just the formula used to come up with it that I am having problems with.
If someone could point me in the right direction I would really appreciate it.

« Last Edit: October 01, 2004, 07:48:00 PM by tammyh75 »


  • Guest
Re:Limiting Reactants
« Reply #1 on: October 01, 2004, 09:57:03 PM »
8.00 moles GeF3H x (3 moles GeF4/4 mole GeF3H) x 100/92.6

Note that since the yield is less than 100%, more moles of reactant are needed than normal by the last factor.
« Last Edit: October 01, 2004, 10:00:57 PM by Demotivator »


  • Guest
Re:Limiting Reactants
« Reply #2 on: October 02, 2004, 04:36:38 PM »
Thank you.  I was trying to make it a lot more complicated than the problem really was.  

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