2 Al(s) + 3 Cl2(g) → 2 AlCl3(s)
Calculate the amount of AlCl3 produced from each reactant and determine the limiting reactant:
mass Al (molar mass) → moles Al(mole-to-mole comparison) → moles AlCl3(molar mass) → mass AlCl3
molar mass Al = 26.982 g/mol
molar mass AlCl3 = 133.332 g/mol
mass AlCl3 = 12.35 g
mass Cl2 (molar mass) → moles Cl2 (mole-to-mole comparison) → moles AlCl3 (molar mass) → mass AlCl3
For the reaction shown, compute the theoretical yield of product (in grams) for each of the following initial amounts of reactants.
2Al(s) + 3Cl2(g) → 2AlCl3(s)
2.5 g Al, 2.5 g Cl2
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