What is the chemical equation for solid aluminum reacts w/ solid iodine to produce solid aluminum iodide?
This is not ideal but we do this.
2Al(s) + 3I2(s) ==> 2AlI3(s)
2Al (s)+ 3I2 (s) >>> 2AlI3 (s) is the normal way we do it. It means this
2Al (s) + 3I2 (s)>>2AlI3
and add the (s) after the aluminum iodide, I forgot.
How would you use the subscripts?
To determine the chemical equation for the reaction between solid aluminum and solid iodine to produce solid aluminum iodide, we need to know the valence of each element involved.
First, let's determine the valence of aluminum. Aluminum is in Group 13 of the periodic table and has an atomic number of 13. It generally loses three electrons to achieve a stable electron configuration, resulting in a +3 charge.
Now, let's determine the valence of iodine. Iodine is in Group 17 of the periodic table and has an atomic number of 53. It generally gains one electron to achieve a stable electron configuration, resulting in a -1 charge.
Considering the valences of aluminum and iodine, we can construct the balanced chemical equation:
2 Al(s) + 3 I2(s) → 2 AlI3(s)
This equation indicates that two moles of solid aluminum react with three moles of solid iodine to produce two moles of solid aluminum iodide. The coefficients (numbers in front of the chemical formulas) ensure that there is an equal number of each atom on both sides of the equation, representing a balanced chemical reaction.