2) Al+Cl_(2)... AlCl_(3) How many grams of aluminum chloride could be produced from 44.0 g of aluminum and 49.0 g of chlorine gas How much of the excess reagent is left? Is this a redox reaction? Show the charges to support your answer.

1. The balanced chemical equation for the reaction is \(2Al + 3Cl_2 \rightarrow 2AlCl_3\). From this equation, 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride. Using stoichiometry, we can determine the amount of aluminum chloride produced from the given amounts of reactants. 2. To determine the amount of the excess reagent left, we need to calculate the amount of each reactant that would be completely consumed in the reaction. The reagent that is not completely consumed is the excess reagent. 3. In this reaction, aluminum loses electrons to form \(Al^{3+}\) and chlorine gains electrons to form \(Cl^-\). Since there is a transfer of electrons, this is a redox reaction.