Answer to Question #306150 in Organic Chemistry for Mum

Question

24/+3Cl₂ → 2AlCl3 In an experiment, 54 grams of aluminium reacted with excess chlorine gas. How many formula units of AICI, are produced?

Answer 1

Solution:

Calculate the moles of aluminium (Al):

The molar mass of Al is 26.98 g/mol

Therefore,

Moles of Al = (54 g Al) × (1 mol Al / 26.98 g Al) = 2.00 mol Al

Balanced chemical equation:

2Al + 3Cl₂ → 2AlCl₃

According to stoichiometry:

2 mol of Al produces 2 mol of AlCl₃

Thus, 2.00 mol of Al produces:

(2.00 mol Al) × (2 mol AlCl₃ / 2 mol Al) = 2.00 mol AlCl₃

One mole of any substance contains 6.022×10²³ formula units.

Therefore,

Formula units of AICI₃ = (2.00 mol AlCl₃) × ( 6.022×10²³ formula units / 1 mol) = 1.2×10²⁴ formula units

Answer: 1.20×10²⁴ formula units of AICI₃ are produced