Balance the chemical equation given below, and determine the number of moles of iodine that reacts with 10.0 g of aluminum. Al(s) + I2(s) → Al2I6(s)

1. 2Al + 3I2 —> Al2I6

2. 0.555mol of I2

Explanation:

1. Al + I2 —> Al2I6

Observing the above equation, there are 2 atoms of Al on the right side and 1 on the left side. To balance it, put 2 in front of Al as shown below:

2Al + I2 —>Al2I6

Also, there are 6 atoms of I on the right side and 2 on the left side. To balance it, put 3 in front I2 as shown below:

2Al + 3I2 —>Al2I6

2. Molar Mass of Al = 27g/mol

Mass of Al = 10g

n = Mass /Molar Mass

n = 10/27 = 0.37mol

From the equation,

2moles of Al reacted with 3 moles of I2.

Therefore, 0.37mol of Al will react with = (0.37 x 3)/2 = 0.555mol of I2

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explanation:

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