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?
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srry. but i don't know
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.
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.
5 answers