Chemistry: 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?

you would need to add or mabye multiply the divide Explanation:...

If 75.0 g silderite ore (feco3) is heated with an

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16.01.2022

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 %

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