Final answer:
To find the mass of I₂ needed to react completely with 35.0 g of Al, we need to balance the chemical equation and use the mole ratio. By calculating the moles of Al and using the mole ratio, we find that 1.297 mol of Al reacts with 1.9465 mol of I₂. Converting the mol to grams, the mass of I₂ needed is 493.36 g.
Step-by-step explanation:
To determine the mass of I₂ needed to react completely with 35.0 g of Al, we first need to balance the chemical equation:
2Al(s) + 3I₂(s) → 2AlI₃(s)
The balanced equation shows that 2 moles of Al react with 3 moles of I₂. To find the moles of I₂, we can use the mole ratio:
Moles of I₂ = Moles of Al x (3 moles I₂ / 2 moles Al)
Given that the molar mass of Al is 26.98 g/mol, we can calculate the moles of Al:
Moles of Al = Mass of Al / Molar mass of Al
Substituting the values, we get:
Moles of Al = 35.0 g Al / 26.98 g/mol Al = 1.297 mol Al
Finally, substituting the moles of Al into the mole ratio equation:
Moles of I₂ = 1.297 mol Al x (3 mol I₂ / 2 mol Al) = 1.9465 mol I₂
To convert the moles of I₂ to grams, we multiply by the molar mass of I₂:
Mass of I₂ = Moles of I₂ x Molar mass of I₂
Using the molar mass of I₂ as 253.80 g/mol, we get:
Mass of I₂ = 1.9465 mol I₂ x 253.80 g/mol I₂ = 493.36 g I₂
Therefore, the mass of I₂ needed to react completely with 35.0 g of Al is 493.36 g.