Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s)  You are given 29.0 g of aluminum and 34.0 g of chlorine gas. a) If you had excess chlorine, how many moles of aluminum chloride could be produced from 29.0 g of aluminum? b) If you had excess aluminum, how many moles of aluminum chloride could be produced from 34.0 g of chlorine gas, Cl2?

Chemistry: An Atoms First Approach
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Chapter5: Stoichiometry
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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 29.0 g of aluminum and 34.0 g of chlorine gas.

a) If you had excess chlorine, how many moles of aluminum chloride could be produced from 29.0 g of aluminum?

b) If you had excess aluminum, how many moles of aluminum chloride could be produced from 34.0 g of chlorine gas, Cl2?

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