Question

When heat is applied to 80 grams of CaCO3, it yields 39 grams of B. Determine the percentage of the yield.

Answer 1

The question is incomplete, the complete question is:

When heat is applied to 80 grams of CaCO3, it yields 39 grams of CaO Determine the percentage of the yield.

CaCO3→CaO + CO2

Answer: The % yield of the product is 87.05 %

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)

We are given:

Given mass of
`CaCO_3` = 80 g

Molar mass of
`CaCO_3` = 100 g/mol

Putting values in equation 1, we get:

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

For the given chemical reaction:

`CaCO_3\rightarrow CaO+CO_2`

By stoichiometry of the reaction:

If 1 mole of
`CaCO_3` produces 1 mole of CaO

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

We know, molar mass of
`CaO` = 56 g/mol

Putting values in above equation, we get:

`\text{Mass of CaO}=(0.8mol* 56g/mol)=44.8g`

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 the product = 39 g

Theoretical value of the product = 44.8 g

Plugging values in equation 2:

`\% \text{yield}=(39 g)/(44.8g)* 100\\\\\% \text{yield}=87.05\%`

Hence, the % yield of the product is 87.05 %