Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s) + 3Cl2(g) 2AlCl3 (s) You are given 25.0 g of aluminum and 30.0 g of chlorine gas ▾ Part A If you had excess chlorine, how many moles of of aluminum chloride could be produced from 25.0 g of aluminum?

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