You are interested in reacting aluminum with iodine to form aluminum iodide. Prior to calculating anything you want to know the limiting reactants. When you have 120 mol Al and 240 mol Iodine, the limiting reagent is ___. However, it was wrong that instead of gm you thought it was mol. Hence, when 120 gm Al and 240 gm iodine are present, the limiting reagent is ___.

When you have 120 mol Al and 240 mol Iodine, the limiting reagent is Al.

When 120 gm Al and 240 gm iodine are present, the limiting reagent is I₂

Explanation:

Determine the reaction:

2Al + 3I₂ → 2AlI₃

2 moles of aluminum react with 3 mol of iodine to produce 2 moles of AlI₃

We can make a rule of three:

2 moles of Al need 3 moles of Iodine to react

Then, 120 moles of Al may need (120 . 3)/2 = 180 moles of I₂

It's ok, because we have 240 moles, so the Iodine is in excess, and the limiting is the Al. Let's confirm it:

3 mol of iodine need 2 moles of Al to react

Then, 240 moles of I, may need (240 . 2)/3 = 160 moles of Al

We do not have enough Al, we need 160 moles and we only have 120.

When we have the mass of each reactant, we need to convert it to moles:

120 g / 26.98 g/mol = 4.44 moles of Al

240 g / 253.8 g/mol = 0.945 moles of I₂

3 mol of iodine need 2 moles of Al to react

Then 0.945 moles of I₂ may need (0.945 .2) / 3 = 0.630 moles of Al

We have 4.44 moles of Al, we need 0.630. Clearly, the Al is the reagent in excess. The limiting reactant is the I₂ but let's verify:

2 mol of Al need 3 moles of I₂ to react

Then, 4.44 moles of Al will need (4.44 . 3) / 2 = 6.66 moles.

We don't have enough iodine, because we only have 0.945 moles

"the periodic table is arranged in order of atomic number."