Question

When limestone rock, which is principally calcium carbonate, is heated, a reaction occurs. If 11.7 g of carbon dioxide were produced in the lab from the decomposition of 30.7g of calcium carbonate, what is the percent yield for the reaction?

Answer 1

Answer: 86.7 %

Step-by-step explanation:

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}`

a) moles of
`CaCO_3`

`\text{Number of moles}=(30.7g)/(100g/mol)=0.307moles`

`CaCO_3\rightarrow CaO+CO_2`

According to stoichiometry :

1 moles of
`CaCO_3` give = 1 mole of
`CO_2`

Thus 0.307 moles
`CaCO_3` give =
`(1)/(1)* 0.307=0.307moles` of
`CO_2`

Mass of
`CO_2=moles* {\text {Molar mass}}=0.307moles* 44g/mol=13.5g`

`\%\text{ yield }=\frac{\text{Actual yield}}{\text{Theoretical yield }}* 100=(11.7g)/(13.5g)* 100=86.7\%`

Therefore, the percent yield for the reaction is, 86.7%