Moles of LiH & LiAlH4: Calculations and Solutions

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SUMMARY

The discussion focuses on the stoichiometric calculations involving lithium hydride (LiH) and lithium aluminum hydride (LiAlH4). It is established that 4 moles of LiH are required to produce 1 mole of LiAlH4, confirming a 4:1 molar ratio. Additionally, the reaction involves aluminum chloride (AlCl3) as a reactant. The balanced chemical equation is LiH + AlCl3 → LiAlH4 + HCl, clarifying the missing components in the initial query.

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  • Understanding of stoichiometry and molar ratios
  • Familiarity with chemical reaction balancing
  • Knowledge of lithium compounds, specifically LiH and LiAlH4
  • Basic chemistry concepts regarding reactants and products
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  • Study the principles of stoichiometry in chemical reactions
  • Learn how to balance chemical equations effectively
  • Research the properties and applications of lithium aluminum hydride (LiAlH4)
  • Explore the role of aluminum chloride (AlCl3) in organic synthesis
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Chemistry students, educators, and professionals involved in inorganic chemistry and reaction mechanisms will benefit from this discussion.

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Homework Statement





*How many moles of LiH are required to make 1 mole of LiAlH4?


* 6 moles of LiH will produce how many moles of LiAlH4?


Can someone help me get started on these?
 
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You start by writing the balancing the relevant equation. Can you do that for the above reaction?
 
I am learning the same stuff in class, I can clearly see from what is written that there is a 4 to 1 ratio, so I think we get 1.5 moles. What I don't see is how the equation works. I wrote it out and somethings seems missing.
LiH + Alxx ----> LiAlH4 + xx
I got myself a bit confused, should there be something for xx or is it just Al?

P.S. I would not normally encroach on someone elses question, but since two weeks have passed, I figured it would be ok.
 
Last edited:
Good thinking. Alxx is actually AlCl_3.
 

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