Aluminum reacts with chlorine gas to form aluminum chloride via the following re

Question

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s)+3Cl2(g)2AlCl3(s)

You are given 19.0 g of aluminum and 24.0 g of chlorine gas.

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 19.0 g of aluminum? Express your answer to three significant figures and include the appropriate units.

If you had excess aluminum, how many moles of aluminum chloride could be produced from 24.0 g of chlorine gas, Cl2? Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

a) The important amount of reactant is aluminium in this case...

19g Al / 26.981g/mol = 0.7042moles (2moles AlCl3 / 2moles Al) = 0.7042moles AlCl3 (133.34g/mol)= 93.897g of AlCl3

b) The important amount of reactant is Chlorine gas

24g Cl2 / 70.906g/mol = 0.3385moles(2moles AlCl3 / 3moles Cl2) = 0.2257moles AlCl3(133.34g/mol) = 30.088g AlCl3

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