Respuesta :
Answer : The theoretical yield of copper is 1.98 grams.
Solution : Given,
Mass of [tex]CuCl_2[/tex] = 2.80 g
Mass of [tex]Al[/tex] = 0.400 g
Molar mass of [tex]CuCl_2[/tex] = 134.45 g/mole
Molar mass of [tex]Al[/tex] = 26.9 g/mole
Molar mass of Cu = 63.5 g/mole
First we have to calculate the moles of [tex]CuCl_2[/tex] and [tex]Al[/tex].
[tex]\text{ Moles of }CuCl_2=\frac{\text{ Mass of }CuCl_2}{\text{ Molar mass of }CuCl_2}=\frac{2.80g}{134.45g/mole}=0.0208moles[/tex]
[tex]\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{0.400g}{26.9g/mole}=0.0149moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]3CuCl_2+2Al\rightarrow 3Cu+2AlCl_3[/tex]
From the balanced reaction we conclude that
As, 3 mole of [tex]CuCl_2[/tex] react with 2 mole of [tex]Al[/tex]
So, 0.0208 moles of [tex]CuCl_2[/tex] react with [tex]\frac{0.0208}{3}\times 2=0.0139[/tex] moles of [tex]Al[/tex]
From this we conclude that, [tex]Al[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]CuCl_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]Cu[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]CuCl_2[/tex] react to give 3 mole of [tex]Cu[/tex]
So, 0.0208 moles of [tex]CuCl_2[/tex] react to give [tex]\frac{0.0208}{2}\times 3=0.0312[/tex] moles of [tex]Cu[/tex]
Now we have to calculate the mass of [tex]Cu[/tex]
[tex]\text{ Mass of }Cu=\text{ Moles of }Cu\times \text{ Molar mass of }Cu[/tex]
[tex]\text{ Mass of }Cu=(0.0312moles)\times (63.5g/mole)=1.98g[/tex]
Therefore, the theoretical yield of copper is 1.98 grams.
