Question

CaCo3 decomposes to give CaO abd Co2 if the masses of CaO and CO2 producef are 5.6g and 4.4g respectively by heating 12g of an impure CaCo3 sample than the percentage impurity of sample will be

Answer 1

Mass of reactant, CaCO  (impure) = 12 g
Mass of Products: CaO = 5.6 g
                               CO = 4.4 g
 According to the Law of Conservation of Mass 
 Mass of Reactants = Mass of Products
 
So,  CaCO = CaO + CO 12 g = (5.6 + 4.4) g

But, 12 g ≠10 g

Therefore, the % Purity of  the sample = (10/12 ) x 100                                                                                                  = 0.83333 x 100 = 83.33%

 Thus, the % Impurity of the sample = (100 - 83.33)% = 16.67% 

Answer 2

Answer : 16.67g

Explanation:

CaCO3 ➡ CaO + CO2

12g 5.6 4.4g

5.6+4.4=10g

We know that when impure substance is converted into its products... The mass obtained is pure not impure .....so using this concept we cn move forward......

OR we can use the concept of law of conservation of mass...

%age purity = pure mass/ total impure mass*100

10/12*100 = 83.33%

Now, this is %purity but we need to find %impurity......

Therefore, (100-83.33)% = 16.67%

Hope it will help..... Pls click on thanks....

also mrk as brainliest... ✨