cerickson248
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When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed? 2Al + 6HCl → 2AlCl3 + 3H2

Respuesta :

happilittlebunni

Well, we need to find the ratio of Al to the other reactant.


Al:HCl = 1:3


--> this means that for every 1 Al used, you have to use 3 HCl.



6*3 = 18 moles of HCl needed to fully react with 6 moles of Al. Since 13<18, HCL is the limiting reactant.



The ratio of HCl:AlCl = 3:1



13/3 = 4.3333...



The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced.



Hope this helps!!! :)


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