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The mass of caco3 produced when co2 gas is bubbled through 500ml of 0.5M ca(oh)2 will be.(please give short detail also)

6 years ago


Answers : (3)


Dear Sunny,


Ca(OH)2 + CO2 = CaCO3 + H2O

0.5 = mass/500*10-3*100

so mass = 25


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Askiitians Expert

Anil Pannikar

IIT Bombay

6 years ago

equivalent wt. of Ca(OH)2 is 36.05 g.

36.05 g of Ca(OH)2  in 1000cm3 (1000 ml) of water gives 1 N solution.

36.05/2 = 18.025 g of Ca(OH)2 in 1000 ml of water gives 0.5 N solution.

so 18.025/2 = 9.0125 g of Ca(OH)2  in 500 ml of water gives 0.5 N solution.

hence, 500 ml solution (0.5 N) contains 9.0125 g of Ca(OH)2.....


Ca(OH) + CO2 ----> CaCO3 + H2O....


molecular mass of Ca(OH)2  is 72.1

" " of CaCO3 is 100.09

Mass of Ca(OH)2 in 500 ml of water is 9.0125 g

Mass of CaCO3 produced is 9.0125x100.09/72 = 12.517 g (approx).


plz approve this answer....

6 years ago

Molarity = no, of moles / Volume of solution [in L]

0.5 M = x moles / 0.5

No. of moles of Ca(OH)2 = 0.25


Ca(OH)2 + CO2 --> CaCO3 + H20

According to the balanced stoichiometric equation, 1 mole of Ca(OH)2 gives 1 mole of CaCO3.

0.25 moles of Ca(OH)2 will give 0.25 moles of CaCO3.

0.25 moles of CaCO3 = .25 * 100 = 25 grams


Hence, 25 grams of CaCO3 is produced.

6 years ago

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