Question

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) What is the maximum mass of aluminum chloride that can be formed when reacting 12.0 g of aluminum with 17.0 g of chlorine? Express your answer to three significant figures and include the appropriate units.

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Answer 1

2 Al + 3 Cl2 = 2 AlCl3

Mole of Al = mass / molar mass

= 12 / 27

= 0.44

Moles of Cl2 = 17.0 / 71

= 0.24

Moles of Chlorine according to Al = 3/2 * mole of Al

( Look Balance reaction)

= 3/2 * 0.44 = 0.66

But we have only 0.24 moles of Cl so Chlorine is limiting reagent

Now moles of AlCl3 = 2/3 mole of Cl

= 2/3 * 0.24

= 0.16

Mass of AlCl3 = mole * molar mass

= 0.16 * 133

= 21.28 grams or 21.3 grams

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