2Al + 3CuCl2.2H2O ==> 3Cu + 2AlCl3 + 2H2O
mols Al = 0.500/27 = 0.0185
mols CuCl2.2H2O = 3.50/170 = 0.0205
mols Cu produced from 0.0185 mols Al could be 0.0185 x (3 mols Cu/2 mols Al) = 0.065
mols Cu produced from 0.0205 mols CuCl2.2H2O could be 0.0205 x (mol Cu/1 mol CuCl2.2H2O) = .0205
In limiting reagent problems you can't produce more than the smallest amount; therefore, CuCl2.2H2O is the LR.
Check all of these calculations.
What is the limiting reactant if 0.500 g Aluminum is reacted with 3.500 g Copper (II) chloride (when calculating the molar mass of copper chloride you need to add mass of 2H2O because it is a dihydrate)?
1 answer
1 answer