Respuesta :
Here we have to get the theoretical yield of Copper in the given reaction.
The theoretical yield of copper in this reaction is 1.306g.
The balance reaction between Aluminium (Al) and CuCl₂.2H₂O (as per the given information it is di-hydrated) to produce copper (Cu) and aluminium trichloride (AlCl₃):
2Al + 3CuCl₂.2H₂O → 3Cu + 2AlCl₃ + 2H₂O↑
The molecular weight of Al, CuCl₂.2H₂O, Cu, AlCl₃ and H₂O are 26.981 , 170.482, 63.546, 133.34 and 18 g/mol respectively.
Now 0.5g Al is equivalent to [tex]\frac{0.5}{26.981}[/tex] = 0.018 moles of Al.
and 3.5 g of CuCl₂.2H₂O is equivalent to [tex]\frac{3.5}{170.482}[/tex] = 0.020 moles of CuCl₂.2H₂O.
Now in this reaction 2 moles of Al reacts with 3 moles of CuCl₂.2H₂O.
Thus 0.018 moles of Al will reacts with [tex]\frac{3}{2}[/tex]×0.018 = 0.027.
But in this reaction only 0.020 moles of CuCl₂.2H₂O is present. So in the reaction all of the CuCl₂.2H₂O will be consumed and it will be the limiting reagent.
So, from (3×170.282) gm of CuCl₂.2H₂O (3×63.546) gm of Cu is produced.
Thus, from 3.5g of CuCl₂.2H₂O
[tex]\frac{190.638}{510.846}[/tex]×3.5 = 1.306g of Cu will produce.
The theoretical yield of copper will be 1.306 grams.
The balanced reaction will be:
[tex]\rm 2\;Al\;+\;3\;CuCl_2.2H_2O\;\rightarrow\;3\;Cu\;+\;2\;AlCl_3\;+\;2\;H_2O[/tex]
Moles = [tex]\rm \frac{weight}{molecular\;weight}[/tex]
Moles of Al in the reaction = [tex]\rm \frac{0.5}{26.981}[/tex]
Moles of Al = 0.018 moles
Moles of [tex]\rm CuCl_2.2H_2O[/tex] in reaction = [tex]\rm \frac{3.5}{170.482}[/tex]
Moles of [tex]\rm CuCl_2.2H_2O[/tex] in reaction = 0.020 moles
From the reaction, 2 moles of Al reacts with 3 moles of [tex]\rm CuCl_2.2H_2O[/tex]
0.018 moles of Al reacts with = [tex]\rm \frac{3}{2}\;\times\;0.018[/tex]
0.018 mole of Al reacts with 0.027 moles of [tex]\rm CuCl_2.2H_2O[/tex]
Since there is only 0.020 moles of [tex]\rm CuCl_2.2H_2O[/tex], thus it is the limiting reactant.
From the reaction,
3 moles of [tex]\rm CuCl_2.2H_2O[/tex] = produces 3 moles of Cu
3 [tex]\times[/tex] 170.482 grams of [tex]\rm CuCl_2.2H_2O[/tex] = 3 [tex]\times[/tex] 63.546 grams of Cu
3.5 grams of [tex]\rm CuCl_2.2H_2O[/tex] = [tex]\rm \frac{3\;\times\;63}{3\;\times\;170}\;\times\;3.5[/tex] grams of Cu
3.5 grams of [tex]\rm CuCl_2.2H_2O[/tex] produces 1.306 grams of Cu.
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