Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)2AlCl3(s) You are given 32.0 g of aluminum and 37.0 g of chlorine gas.

1. If you had excess chlorine, how many moles of of aluminum chloride could be produced from 32.0 g of aluminum? Express your answer numerically in moles.

2. If you had excess aluminum, how many moles of aluminum chloride could be produced from 37.0 g of chlorine gas, Cl2 ? Express your answer numerically in moles.