Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:
2Al(s) + 3Cl₂ (g)→2AlCl3 (s)
You are given 31.0 g of aluminum and 36.0 g of chlorine gas.
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Part A
If you had excess chlorine, how many moles of of aluminum chloride could be produced from 31.0 g of aluminum?
Express your answer to three significant figures and include the appropriate units.

Please help! I cannot solve it!