Question

Quicklime, CaO, can be prepared by roasting limestone, CaCO3, according to the following reaction. CaCO3(s) ∆→ CaO(s) + CO2(g). When 2.00 × 103 g CaCO3 is heated, the actual yield of CaO is 1.05 × 103 g. What is the percentage yield?

Answer 1

The percentage yield of CaO can be calculated using the formula (Actual yield / Theoretical yield) x 100%. In this case, the actual yield is 1.05 × 103 g and the theoretical yield is 1.119 × 103 g, resulting in a percentage yield of 93.8%.

Step-by-step explanation:

The percentage yield can be calculated using the formula:

Percentage yield = (Actual yield / Theoretical yield) x 100%

In this case, the actual yield of CaO is given as 1.05 × 103 g. To calculate the theoretical yield, we need to determine the molar ratio between CaCO3 and CaO. From the balanced equation, we can see that 1 mole of CaCO3 reacts to form 1 mole of CaO. So the molar mass of CaO (56.08 g/mol) is equal to the molar mass of CaCO3 (100.09 g/mol).

Theoretical yield = (2.00 × 103 g CaCO3) x (56.08 g/mol CaO / 100.09 g/mol CaCO3) = 1.119 × 103 g CaO

Now we can calculate the percentage yield:

Percentage yield = (1.05 × 103 g / 1.119 × 103 g) x 100% = 93.8%

Answer 2

93.71 %

Step-by-step explanation:

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

For
`CaCO_3` :-

Mass of
`CaCO_3` =
`2.00* 10^3` g

Molar mass of water = 100.0869 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (2.00* 10^3\ g)/(100.0869\ g/mol)`

`Moles\ of\ CaCO_3= 19.98\ mol`

According to the given reaction:

`CaCO_3_((s))\rightarrow CaO_((s))+CO_2_((g))`

1 mole of
`CaCO_3` on reaction forms 1 mole of
`CaO`

19.98 mole of
`CaCO_3` on reaction forms 19.98 mole of
`CaO`

Moles of
`CaO` = 19.98 moles

Molar mass of
`CaO` = 56.0774 g/mol

Mass of sodium sulfate = Moles × Molar mass = 19.98 × 56.0774 g = 1120.43 g

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

`\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}* 100`

Given , Values from the question:-

Theoretical yield = 1120.43 g

Experimental yield = 1050 g

Applying the values in the above expression as:-

`\%\ yield =(1050)/(1120.43)* 100`

`\%\ yield =93.71\ \%`