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}=(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
`(0.647)/(4)* 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* \text{ Molar mass of }Fe_2O_3`

`\text{ Mass of }Fe_2O_3=(0.324moles)* (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}* 100`

`\% \text{ yield of the reaction}=(45.0g)/(51.7g)* 100=87.0\%`

Therefore, the percent yield of the reaction is, 87.0 %