Answer:
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
