Question

11. When 1.25 g of a chalk sample is strongly heated, 0.44 g CO2 gas is produced. Determine the percentage
of pure CaCo3 in the chalk sample.

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Answer 1

Answer:

MARK ME AS BRAINLIST

Explanation:

Using ideal gas law,

PV=nRT;

Converting P in atm, V in litres and T in K, n=0.01196~ 0.012

1 mole of CaCO3= 1 mole of CO2.

Therefore number of moles of CaCO3=

0.012.gram molecular mass of CaCO3= 0.012x100= 1.196.

% purity=100-( (1.54-1.196)/1.54x100)

= 77.66% purity