The answer is A. 5.33 mol.
How to find moles?
To determine the number of moles of Cl₂ required to form 3.55 mol of AlCl₃, use the stoichiometric coefficients from the balanced chemical equation:
2Al(s) + 3Cl₂(g) ⟶ 2AlCl₃(s)
This equation tells us that 3 moles of Cl₂ are required to react with 2 moles of AlCl₃. Therefore, set up a proportion:
3 moles Cl₂ / 2 moles AlCl₃ = x moles Cl₂ / 3.55 moles AlCl₃
Solving for x:
x = (3 moles Cl₂/2 moles AlCl₃) × (3.55 moles AlCl₃)
x = 5.33 moles Cl₂
Therefore, 5.33 moles of Cl₂ are required to form 3.55 mol of AlCl₃.
Complete question:
Using the following balanced equation, 2al(s) 3cl2(g)⟶2alcl3(s) how many moles of cl2 are required to form 3.55mol alcl3?
A. 5.33 mol
B. 2.67 mol
C. 3.55 mol
D. 7.10 mol