Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.
The balanced chemical equation is AlI₃(s) + 3 K(s) → 3 KI(s) + Al(s).
Given the reaction has a percent yield of 67.8%, what is the mass in grams of aluminum iodide that would be required to yield an actual amount of 30.25 grams of aluminum?
AlI₃(s) + 3 K(s) → 3 KI(s) + Al(s).
You wan 30.25 g Al. The rxn is 67.8% yield; therefore, you need 30. 25/0.678 = 44.6 g Al if the reaction were 100%.
44.6 g Al/27 = approx 1.65 mols Al. You will need 1.65 mol AlI3 since the equation shows 1 mol AlI3 produces 1 mol Al in the balanced equation.
Convert 1.65 mols AlI3 to grams. g = mols x molar mass = ?
Check my calculations. Post your work if you get stuck.
To find the mass of aluminum iodide required to yield 30.25 grams of aluminum, we need to calculate the stoichiometric amount of aluminum iodide used in the reaction.
1. Start with the given mass of aluminum (30.25 grams) and use the molar mass of aluminum (Al) to convert it to moles. The molar mass of aluminum is 26.98 g/mol.
Moles of Aluminum = Mass of Aluminum / Molar Mass of Aluminum
= 30.25 g / 26.98 g/mol
=~ 1.1213 mol (rounded to four decimal places)
2. Now, using the balanced chemical equation, we can determine the mole ratio between aluminum iodide (AlI₃) and aluminum (Al). From the balanced equation, we can see that 1 mole of aluminum reacts with 1 mole of aluminum iodide.
Moles of Aluminum Iodide = Moles of Aluminum
= 1.1213 mol
3. Finally, we convert the moles of aluminum iodide to grams using its molar mass. The molar mass of aluminum iodide (AlI₃) is 407.69 g/mol.
Mass of Aluminum Iodide = Moles of Aluminum Iodide x Molar Mass of Aluminum Iodide
= 1.1213 mol x 407.69 g/mol
=~ 456.91 g (rounded to two decimal places)
Therefore, to yield an actual amount of 30.25 grams of aluminum, approximately 456.91 grams of aluminum iodide would be required.