Question

by accomposition of 50g CaCo3 , 11g CO2 is obtained then find % purity of CaCo3​

Answer 1

Answer:

% purity = 25/50*10 = 50 %

Explanation:

CaCo3 _ Cao +Co2

moles of CaCo3 = 0.25 (stoichiometry ratio 1:1)

Mass = 100* 0.25 = 25 gram of pure CaCo3

% purity = 25/50*10 = 50 %