Limiting Reactant and Percent Yield Answers

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0.50 g Al equals moles

0.25 g Al equals moles

Compare this ratio to the experimental number of moles computed above. The limiting reactant(s) is/are .

Convert mass to moles for both reactants. (round to 2 significant figures.) 2.50 g CuCl2 equals moles

3CuCl2 + 2Al 2AlCl3 + 3Cu Based on the balanced equation, the molar ratio of CuCl2 to Al should be .

Analysis: The limiting reactant(s) appeared to be because . . .

Subtract the mass of the filter paper (0.27 g) from the mass of the paper and copper (0.98 g). Record the difference in the data table as the amount of copper.

You already converted mass to moles for the reactants. Now convert mass to moles for the product, copper.0.71 g Cu equals moles

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Convert mass to moles for both reactants.(round to 2 significant figures.) 2.50 g CuCl2 equals moles

3CuCl2 + 2Al 2AlCl3 + 3Cu Based on the balanced equation, the molar ratio of CuCl2 to Al should be 3:2. Compare this ratio to the experimental number of moles computed above. The limiting reactant(s) is/are .

3CuCl2 + 2Al 2AlCl3 + 3Cu Using the molar ratio between the limiting reactant (aluminum) and the product copper, determine the theoretical yield of copper.

Determine the percent yield of copper by dividing the actual yield (in moles) by the theoretical yield (in moles).