Answer : The percent yield of the reaction is, 87.0 %
Solution : Given,
Mass of \(FeCO_3\) = 75.0 g
Molar mass of \(FeCO_3\) = 115.85 g/mole
Molar mass of \(Fe_2O_3\) = 159.69 g/mole
First we have to calculate the moles of \(FeCO_3\).
\(\text{ Moles of }FeCO_3=\frac{\text{ Mass of }FeCO_3}{\text{ Molar mass of }FeCO_3}=\frac{75.0g}{115.85g/mole}=0.647moles\)
Now we have to calculate the moles of \(Fe_2O_3\)
The balanced chemical reaction is,
\(4FeCO_3(s)+O_2(g)\rightarrow 2Fe_2O_3(s)+4CO_2(g)\)
From the balanced reaction we conclude that
As, 4 mole of \(FeCO_3\) react to give 2 mole of \(Fe_2O_3\)
So, 0.647 moles of \(FeCO_3\) react to give \(\frac{0.647}{4}\times 2=0.324\) moles of \(Fe_2O_3\)
Now we have to calculate the mass of \(Fe_2O_3\)
\(\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3\)
\(\text{ Mass of }Fe_2O_3=(0.324moles)\times (159.69g/mole)=51.7g\)
Theoretical yield of \(Fe_2O_3\) = 51.7 g
Experimental yield of \(Fe_2O_3\) = 45.0 g
Now we have to calculate the percent yield of the reaction.
\(\% \text{ yield of the reaction}=\frac{\text{ Experimental yield of }Fe_2O_3}{\text{ Theretical yield of }Fe_2O_3}\times 100\)
\(\% \text{ yield of the reaction}=\frac{45.0g}{51.7g}\times 100=87.0\%\)
Therefore, the percent yield of the reaction is, 87.0 %
