Chemistry help please! having trouble?

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Exotic_Star

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A chemist mixes 3.02 grams of aluminum metal with 15.1 grams of elemental iodine.


How many moles of each reactant were mixed? Show Calculations.

A. Moles elemental Aluminum


B. Moles elemental Iodine.

Demonstrate clearly which reactant is the limiting reagent.
 
Atomic wqeights: Al=27, I=127, I2=254, AlI3=408

2Al + 3I2 ===> 2AlI3

A. 3.02gAl x 1molAl/27gAl = 0.112 mole Al

B. 15.1gI2 x 1molI2/254gI2 = 0.059 mole I2

From the balanced equation, 3 moles of I2 needs 2 moles Al. Or, 3/2 mole I2 needs 1 mole Al. So 0.112 mole Al needs (3/2)(0.112) = 0.168 mole I2. You don't have that much I2, so I2 is the limiting reagent: It will run out first.
 
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