10.0 g of aluminium react with 35.0g of chlorine gas to produce aluminium chloride. Which reactant is limiting, which is in excess, and how much product is produced?
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2Al + 3Cl2 = 2AlCl3
54g 213 g 267 g
213 g of chlorine needs 54 g of Al
35 g of chlorine needs 54 x 35 / 213 = 8.87 g Al.
It means chlorine will be completely used up in the reaction. hence chlorine is the limiting reagent.
aluminium is the excess reagent as 10-8.87 = 1.13 g remains.
Product formed = 8.87 + 35 = 43.87 g of AlCl3
Input = 10+35 ( Al + Cl2) = 45
Output = 1.13 + 43.87 (Al + AlCl3) = 45
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