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Problem- Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)----=2AlCl3(s) You are given 31.0g of aluminum and 36.0g of chlorine gas.
1) If you had excess chlorine, how many moles of of aluminum chloride could be produced from 31.0g of aluminum?
If you had excess aluminum, how many moles of aluminum chloride could be produced from 36.0g of chlorine gas, Cl2?
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