use the following balanced chemical equation to calculate: (.?3CuCl2 + 2Al -gt; 3u + 2AlCl3?Mass of beaker:54.31gMass of aluminum foil:0.40gMass of beaker + copper + h2O:54.85g use the following balanced chemical equation to calculate:(.)a) IDENTIFY THE LIMITING REAGENT AND THE EXCESS REAGENT IN THIS REACTION.wHAT VISIBLE EVIDENCE IS THERE TO CONFIRM YOUR IDENTIFICATION?for this one is it something like that?Firstly, we would convert the grams of each reactant to moles using: Moles mass (g)/ molar mass (g/mol): M Cu 63.55, M CI 35.45270.9, M CuCI2 134.45g/mol Moles CuCI2 45.85g/ 134.45g/mol 0.339 moles (in reaction mixture) Moles Al 0.40g/ 26.98g/mol 0.015 moles (in reaction mixture)Dividing moles of each reactant by its coefficient: CuCI2 134.45g/moles/ 3 44.817Al 0.015 moles/ 2 0.008CuCI2 has the larger number so it is present in excessis that truee??im confused and it due tomorrow!!;(and what about the percentage yield of copper ??
Sorry that I did not see this question any earlierYour data is confusingFirst, is the reaction a solution of Copper (II) chloride or solid CuCl2? Should the balanced equation be: 3 CuCl2 (aq) + 2 Al (s) - 3 Cu (s) + 2 AlCl3 (aq)? Mass of beaker + copper + h2O:54.85g Where did the water come from? If it is from the aqueous solution, why was the metallic Copper not dried? Exactly how much CuCl2 was used (or how much of what concentration of CuCl2 solution was used)? 45.85 g of CuCl2 or of solution? You have the right idea of how to calculate the number of moles of the reactantsIf all of the Aluminum foil reacts, it was not in excess (so the CuCl2 must be in excess)For every 2 moles of Al that reacts, you should get 3 moles of Cu metal0.40 g / 26.98 g/mol 0.0148 moles of AlThis should make 0.0222 moles of Cu or 0.0222 63.55 g/mol 1.41 grams of Copper as the theoretical yield.
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