if you react 5.00g of aluminum with 20.0g of iodine, how much aluminum iodide do you expect to produce from this reaction? what is the limiting reactant ? what is the excess reactant? how much of the excess reactant do you expect to have left over?
2Al + 3I2 >>>2AlI3
so for each two moles of aluminum, one needs three moles of I2.
you have 5/27= .185 moles of Al
so you need then 3/2 *.185 moles of I2, or
1.5*.185*127*2=70.5 grams. You have far less than that, so Iodine is the limiting reageant.
How much can you expedt:
moles of I2=20/(2*127)=.079 moles, so you should get 2/3*.079 moles AlI3.
To determine the amount of aluminum iodide that can be produced, we first need to find the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
Step 1: Write a balanced chemical equation for the reaction:
2Al + 3I2 → 2AlI3
From the balanced equation, we can see that 2 moles of aluminum react with 3 moles of iodine to produce 2 moles of aluminum iodide.
Step 2: Convert the given masses of reactants to moles:
The molar mass of aluminum (Al) is 26.98 g/mol.
The molar mass of iodine (I2) is 253.8 g/mol.
Mass of aluminum (Al) = 5.00 g
Moles of aluminum (Al) = Mass / Molar mass = 5.00 g / 26.98 g/mol = 0.185 mol
Mass of iodine (I2) = 20.0 g
Moles of iodine (I2) = Mass / Molar mass = 20.0 g / 253.8 g/mol = 0.079 mol
Step 3: Determine the stoichiometric ratio:
From the balanced equation, we can see that the ratio of moles of aluminum to moles of iodine is 2:3.
If 2 moles of aluminum correspond to 3 moles of iodine, we can set up a ratio:
0.185 mol of aluminum / 2 mol of aluminum = 0.093 mol of iodine / 3 mol of iodine
Step 4: Identify the limiting reactant:
The limiting reactant will be the one that produces fewer moles of the product. From the calculated ratio above, we can see that for the given amounts of reactants, the iodine is the limiting reactant since it would produce fewer moles of aluminum iodide.
Step 5: Calculate the moles of aluminum iodide formed:
From the stoichiometric ratio, we know that for every 3 moles of iodine, 2 moles of aluminum iodide are produced.
Moles of iodine (I2) = 0.079 mol (limiting reactant)
Moles of aluminum iodide (AlI3) = (2/3) * Moles of iodine = (2/3) * 0.079 mol ≈ 0.053 mol
Step 6: Calculate the mass of aluminum iodide:
Molar mass of aluminum iodide (AlI3) = 407.7 g/mol
Mass of aluminum iodide (AlI3) = Moles * Molar mass = 0.053 mol * 407.7 g/mol ≈ 21.59 g
Therefore, you can expect to produce approximately 21.59 g of aluminum iodide from the given reactants.
The limiting reactant is iodine (I2).
To find the excess reactant and the amount left over, we will calculate the amount of the excess reactant left after the reaction with the limiting reactant.
Step 7: Calculate the reaction that occurs between the limiting reactant and excess reactant:
From the balanced equation, we know that 2 moles of aluminum react with 3 moles of iodine to produce 2 moles of aluminum iodide.
Using the stoichiometry, we can find the moles of iodine (excess reactant) needed to react with the given moles of aluminum:
Moles of aluminum (Al) = 0.185 mol
Moles of iodine required = (3/2) * Moles of aluminum = (3/2) * 0.185 mol ≈ 0.278 mol
Step 8: Calculate the excess reactant remaining:
Moles of iodine initially = 0.079 mol
Moles of iodine required (from step 7) = 0.278 mol
Excess moles of iodine = Moles initially - Moles required = 0.079 mol - 0.278 mol = -0.199 mol
(Note: A negative value indicates that the iodine is in excess.)
Step 9: Calculate the mass of excess reactant remaining:
Mass of iodine (excess reactant) = Moles * Molar mass = -0.199 mol * 253.8 g/mol ≈ -50.63 g
Since the mass cannot be negative, there will be no excess iodine remaining.
In summary:
- You can expect to produce approximately 21.59 g of aluminum iodide.
- The limiting reactant is iodine (I2).
- The excess reactant is aluminum (Al).
- There will be no excess reactant left over.