(Solved): 2Al(s)+3CuCl2(aq)2AlCl3(aq)+3Cu(s) A 5.396g piece of aluminum is cut from an aluminum pi ...

$2\mathrm{Al}(s)+3{\mathrm{CuCl}}_2(aq)\to 2{\mathrm{AlCl}}_3(aq)+3\mathrm{Cu}(s)$ A $5.396\mathrm{g}$ piece of aluminum is cut from an aluminum pie pan. It is submersed in an aqueous solution of $53.78\mathrm{g}$ of copper(II)chloride in $600.0\mathrm{mL}$ of solution and reacts as shown above. (a) What is the limiting reagent? (b) How many grams of the excess reagent will remain at the end of the reaction? 9 (c) How many moles of the excess reagent are used up in the reaction? $\mathrm{mol}$ (d) What mass of copper will be produced from this reaction? 9 (e) Will copper be visible in the reaction vessel at the end of this reaction? (f) Will aluminum be visible in the reaction vessel at the end of this reaction?