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Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s) 3Cl2(g)→2AlCl3(s) You are given 28.0 g of aluminum and 33.0 g of chlorine gas.
If you had excess chlorine, how many moles of of aluminum chloride could be produced from 28.0 g of aluminum?

Sagot :

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