Solving a Stochiometry Question: 2Li + F2 --> 2LiF

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The discussion centers on the stoichiometry of the reaction between lithium and fluorine to produce lithium fluoride (2Li + F2 --> 2LiF). Participants emphasize the importance of identifying the limiting reactant, which is crucial for determining the amount of lithium fluoride produced. The correct molar mass calculations are clarified, noting that fluorine exists as diatomic molecules (F2), while lithium is monoatomic. Proper forum etiquette is also highlighted, suggesting that such questions should be posted in the Homework & Coursework section.

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A 15-g sample of lithium is reacted with 15 g of fluorine to form lithium fluoride: 2Li + F2 --> 2LiF. After the reaction is complete, what will be present?
I think I start by dividing 15 g by the molar masses, but is the molar mass
2 x amu of Li and 2 x amu of F?
Thanks
 
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help is hereeeee...

i think you have to first find out what the limiting reactant is. and then use that amount to see how much lithium fluoride forms.
hope this helps :biggrin:
 
SCGirl said:
A 15-g sample of lithium is reacted with 15 g of fluorine to form lithium fluoride: 2Li + F2 --> 2LiF. After the reaction is complete, what will be present?
I think I start by dividing 15 g by the molar masses, but is the molar mass
2 x amu of Li and 2 x amu of F?
Thanks
Correct for fluorine, because it is made up of diatomic molecules (hence, F2), but not lithium, which is monoatomic.

PS: From the next time on, such questions should be posted under the Homework & Coursework forums found near the top of the main page.
 

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