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Aluminum metal and iodine gas react to produce aluminum chloride: Al+ I2 AlI3 A a. If 1.20 g of Al and 2.40 g of iodine

Aluminum metal and iodine gas react to produce aluminum chloride:

Al+ I2→ AlI3
A a. If 1.20 g of Al and 2.40 g of iodine are allowed to react, how many grams of aluminum iodides can be made? Hint: Identify the limiting reagent.

How many moles of excess reagent remains after the reaction has finished?

If 1.29 g are obtained from the reaction in part a, what is the percent yield of this reaction?





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