Question

50 g of a sample of limestone (CaCO3) on complete
decomposition gives 20 g of CO2. The percentage
purity of CaCO3 in limestone is
(Atomic mass of Ca = 40 u)
(1) 75%
(2) 85%
(3) 95.2%
(4) 90.9%​

Answer 1

Answer:

answer is 90.9%

Explanation:

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Answer 2

Answer:

90.9%.

Explanation:

From the question we get that for 100 g of the CaCO3 we will have 44g of the CO2. Hence, for 50 g of the CaCO3 we will have 50/100*44 which is 22g. Now, for 1 g of the CO2 we will have 50/22 g of CaCO3 hence, for 20 g of the CO2 20*50/22 g of CaCO3 which is 45.4 g of CaCO3 and this is the pure limestone.

So, the percentage purity of the CaCO3 will be 45.4/50 *100 which on solving we will get approximately 90.9 % pure CaCO3.