Question

12 grams of a mixture of sand and calcium carbonate on strong heating produced 7.6 grams of residue. How many grams of sand is present in the mixture ?​

Answer 1

Answer:2.38g

Explanation:

From the balanced equation: CaCO3(s)=CaO(s) + CO2(g)

Number of moles of CaO = $\frac{7.6}{56}$

= 0.1357 moles

Using mole ratio; CaCO3 :CaO

Moles of CaCO3 = $\frac{1}{1}$×0.1357moles

=0.1357moles

We can now calculate the mass of CaCO3 using the number of mole available.

Mass of CaCO3 = 0.1357×106

=14.38g of CaCO3

Mass of sand present

14.38g-12.0g

=2.38g

Answer 2

Given:

Total mass = 12 gm

Mass of residue = 7.6 gm

To Find:

The mass of sand present in the mixture.

Calculation:

- Sand is neither volatile nor it decomposes on heating.

- Only calcium carbonate undergoes thermal decomposition to evolve carbon dioxide gas and calcium oxide as per the reaction:

CaCO3 → CaO + CO2

⇒ Only CO2 is evolved in the heating of the given mixture.

⇒ The mass of CO2 evolved = 12 - 7.6 = 4.4 gm

- According to the given reaction, 44 gm of CO2 is evolved by 100 gm of CaCO3.

⇒ 4.4 gm of CO2 is evolved by 10 gm of CaCO3.

- So, the mass of sand in the mixture = 12 - 10 = 2 gm

- So, the original mixture contains 2 grams of sand in it.