Question

if 75.0 g silderite ore (FeCO3) is heated with an excess of oxygen, 45.0 g of ferric oxide (Fe2O3) is produced. 4FeCO3(s) + O2(g) → 2Fe2O3(s) + 4CO2(g) What is the percent yield of this reaction?

Answer 1

Hope this helps you.

Answer 2

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 %