Question

25 g of 80% pure Caco3, sample was decomposed
completely. Amount of Cao formed is
(1) 14 g
(2) 11. 2 g
(3) 20 g
(4) 8 g


Correct ans is 11.2 gm, pls give the correct explanation for it!! ​

Answer 1

Answer:

Explanation:

I think this image will give ur answer

Answer 2

Answer:

Explanation:

The mass of pure Calcium carbonate ( CaCO3) is :

25/100 × 80 = 20g

Equation for the reaction is :

CaCO₃ (s) + 2HCl (aq) —> CaCl₂(s) + CO₂(g) + H₂O(l)

The mole ratio of Calcium carbonate and the CO₂ liberated will be 1 : 1

Moles of Calcium carbonate = mass of calcium carbonate / molar mass

The Molar mass of Calcium carbonate = 100

Mole = 20/100 = 0.2 moles

Moles of CO₂ = 0.2 moles

Molar gas volume at NTP is :

1 mol = 22.4 litres

0.2 moles =?

0.2/ 22.4 = 11.2 litres

CO₂ liberated is 11.2  litres