Answer : The percent yield of the reaction is, 86.3 %
Solution : Given,
Mass of \(CuCl_2\) = 15 g
Mass of \(NaNO_3\) = 20 g
Molar mass of \(CuCl_2\) = 134.45 g/mole
Molar mass of \(NaNO_3\) = 84.9 g/mole
Molar mass of \(NaCl\) = 58.44 g/mole
First we have to calculate the moles of \(CuCl_2\) and \(NaNO_3\).
\(\text{ Moles of }CuCl_2=\frac{\text{ Mass of }CuCl_2}{\text{ Molar mass of }CuCl_2}=\frac{15g}{134.45g/mole}=0.112moles\)
\(\text{ Moles of }NaNO_3=\frac{\text{ Mass of }NaNO_3}{\text{ Molar mass of }NaNO_3}=\frac{20g}{84.9g/mole}=0.236moles\)
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
\(CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2\)
From the balanced reaction we conclude that
As, 1 mole of \(CuCl_2\) react with 2 mole of \(NaNO_3\)
So, 0.112 moles of \(CuCl_2\) react with \(0.112\times 2=0.224\) moles of \(NaNO_3\)
From this we conclude that, \(NaNO_3\) is an excess reagent because the given moles are greater than the required moles and \(CuCl_2\) is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of \(NaCl\)
From the reaction, we conclude that
As, 1 mole of \(CuCl_2\) react to give 2 mole of \(NaCl\)
So, 0.112 moles of \(CuCl_2\) react to give \(0.112\times 2=0.224\) moles of \(NaCl\)
Now we have to calculate the mass of \(NaCl\)
\(\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl\)
\(\text{ Mass of }NaCl=(0.224moles)\times (58.44g/mole)=13.09g\)
Theoretical yield of \(NaCl\) = 13.09 g
Experimental yield of \(NaCl\) = 11.3 g
Now we have to calculate the percent yield of \(NaCl\)
\(\% \text{ yield of }NaCl=\frac{\text{ Experimental yield of }NaCl}{\text{ Theretical yield of }NaCl}\times 100\)
\(\% \text{ yield of }NaCl=\frac{11.3g}{13.09g}\times 100=86.3\%\)
Therefore, the percent yield of the reaction is, 86.3 %
