Aluminum chloride is used as a catalyst in various industrial reactions. It is prepared from hydrogen chloride gas and solid aluminum metal shavings. Hydrogen gas is the other product.

HCl(g) + Al(s) yields AlCl3(s) + H2(g)

Suppose a reaction vessel contains 13.82 grams of Al (s) and 27.9g of HCl(g). How many grams of AlCl3 can be prepared from this mixture?

Question

Aluminum chloride is used as a catalyst in various industrial reactions.  It is prepared from hydrogen chloride gas and solid aluminum metal shavings.  Hydrogen gas is the other product. 

            

HCl(g)   +    Al(s)          yields    AlCl3(s)  +    H2(g)

 

 

Suppose a reaction vessel contains 13.82 grams of Al (s) and 27.9g of HCl(g).  How many grams of AlCl3 can be prepared from this mixture?

anonymous · Answer

really confused on how to solve this

koikkara · Answer

umm...usually the limiting reagent in the hydrogen chloride. 27.9g of HCl contains 27.1g of chlorine. 
Assuming 100% consumption, this will make 34g of AlCl3, leaving 6.9g of Al unreacted. 

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koikkara · Answer

@sat_chen So whats the answer u  needed ?

anonymous · Answer

um u just copied an answer off yahoo answers thanks i can google too. I want someone to explain it better to me and also it doesnt feel right no steps shown at all

gebooors · Answer

Balanced equation is 6 HCL + 4 Al - > 2 AlCl3 + 3 H2
Calculate amounts of substances,
Is n(HCl) more or less than 1,5 times n(Al)?
You find out limiting reactant.