Question

When heat is applied to 80 grams of CaCO3, it yields 39 grams of CO2. Determine

the percentage of the yield.

Answer 1

Answer: The percentage yield of
`CO_2` is 90.26%.

Step-by-step explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` ......(1)

Given mass of
`CaCO_3` = 80 g

Molar mass of
`CaCO_3` = 100 g/mol

Plugging values in equation 1:

`\text{Moles of }CaCO_3=(80g)/(100g/mol)=0.8 mol`

For the given chemical equation:

`CaCO_3\rightarow CaO+CO_2`

By the stoichiometry of the reaction:

If 1 mole of
`CaCO_3` produces 1 mole of
`CO_2`

So, 0.8 moles of
`CaCO_3` will produce =
`(1)/(1)* 0.8=0.8mol` of
`CO_2`

Molar mass of
`CO_2` = 44 g/mol

Plugging values in equation 1:

`\text{Mass of }CO_2=(0.8mol* 44g/mol)=35.2g`

The percent yield of a reaction is calculated by using an equation:

`\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}* 100` ......(2)

Given values:

Actual value of
`CO_2` = 35.2 g

Theoretical value of
`H_2CO_3` = 39 g

Plugging values in equation 2:

`\% \text{yield of }CO_2=(35.2g)/(39g)* 100\\\\\% \text{yield of }CO_2=90.26\%`

Hence, the percentage yield of
`CO_2` is 90.26%.