Question

The decomposition of 3.32 g CaCO3 yields 1.24 g CaO. What is the percent yield of this reaction? (CaCO3 --> CaO + CO2)

Answer 1

66.57%

Step-by-step explanation:

The decomposition reaction of calcium carbonate is shown below as:

`CaCO_3\rightarrow CaO+CO_2`

Calculation of moles of
`CaCO_3` :

Amount = 3.32 g

Molar mass of
`CaCO_3` = 100 g/mol

The formula for the calculation of moles is shown below:

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

Thus, moles are:

`moles= (3.32\ g)/(100\ g/mol)`

`moles= 0.0332\ mol`

According to reaction,

0.0332 moles of
`CaCO_3` decomposes to yield 0.0332 moles of
`CaO`

Theoretical yield = 0.0332 moles

Calculation of moles of
`CaO` formed as:

Amount = 1.24 g

Molar mass of
`CaO` = 56 g/mol

The formula for the calculation of moles is shown below:

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

Thus, moles are:

`moles= (1.24\ g)/(56\ g/mol)`

`moles= 0.0221\ mol`

Experimental yield = 0.0221 moles

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

Thus,

`Yield\%=\frac {0.0221}{0.0332}* 100`

Percent yield = 66.57%