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