How Much Iodine Reacts with 23g of Aluminum?

[mightyhealthy]

Calculate the mass in grams of iodine(I₂) that will react completely with 23g of aluminum (Al) to form aluminum iodide (AlI₃).

 

[Borek]

Starting with reaction equation will definitely help.

 

[Blueshift5]

To calculate the mass of iodine that will react completely with 23g of aluminum, we need to first determine the balanced chemical equation for the reaction. The reaction between aluminum and iodine is represented as:

2Al + 3I2 → 2AlI3

From this equation, we can see that for every 2 moles of aluminum (Al) that react, 3 moles of iodine (I2) are required. This means that the ratio of iodine to aluminum is 3:2.

To calculate the moles of aluminum present in 23g, we can use the molar mass of aluminum, which is 26.98 g/mol. Therefore, 23g of aluminum is equal to 0.85 moles (23g/26.98 g/mol).

Using the ratio of 3 moles of iodine to 2 moles of aluminum, we can calculate the moles of iodine required to react with 0.85 moles of aluminum. This is equal to (3/2) x 0.85 = 1.275 moles of iodine.

Finally, to convert moles to grams, we use the molar mass of iodine, which is 126.90 g/mol. Therefore, the mass of iodine required to react completely with 23g of aluminum is 1.275 moles x 126.90 g/mol = 161.85g.

In conclusion, 161.85g of iodine (I2) will react completely with 23g of aluminum (Al) to form aluminum iodide (AlI3).