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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)
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
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)2 + 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....
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.
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