Question

Number of mole of co2 produced on heating 20 g of caco3(s), given that caco3 is only 75% pure is

Answer 1

Answer:

0.15 moles

Explanation:

since only 75% of CaCO3 is pure

Thus mass of pure CaCO3= 75/100*20 g = 15 g

15 g CaCO3 = 0.15 mole

CaCO3→CO2 + CaO

Since by equation we get that 1 mole of CaCO3 produces 1 mole CO2

Therefore 0.15 moles of CaCO3 will give 0.15 moles CO2.

Answer 2

Answer:

3/20

Explanation:

75×20/100

15/100

3/20