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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)

Respuesta :

Eduard22sly

Answer:

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

dsdrajlin

The balanced chemical equation is:

2 Al(s) + 3 I₂(s) → Al₂I₆(s)

0.557 moles of iodine react with 10.0 g of aluminum.

Let's consider the following unbalanced equation.

Al(s) + I₂(s) → Al₂I₆(s)

We will balance it using the trial and error method. We can get the balanced equation by multiplying Al by 2 (balance Al atoms) and I₂ by 3 (balance I atoms).

2 Al(s) + 3 I₂(s) → Al₂I₆(s)

The molar mass of Al is 26.98 g/mol. The moles corresponding to 10.0 g of Al are:

[tex]10.0 g \times \frac{1mol}{26.98g} = 0.371 mol[/tex]

The molar ratio of Al to I₂ is 2:3. The moles of I₂ that react with 0.371 moles of Al are:

[tex]0.371 molAl \times \frac{3molI_2}{2molAl} = 0.557 mol I_2[/tex]

The balanced chemical equation is:

2 Al(s) + 3 I₂(s) → Al₂I₆(s)

0.557 moles of iodine react with 10.0 g of aluminum.

You can learn more about stoichiometry here: https://brainly.com/question/9743981

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