What will be the amount of CaCO3 on being passed through 5.6 cubic decimeter of CO2 at NTP through excess of slaked lime [Ca(OH)2].

Asked by mesamit | 31st Jul, 2010, 03:41: PM

Expert Answer:

 

Ca(OH)2 + CO2 ---> CaCO3 +H2O

Since the equation is balanced it means that 1 mole of carbondioxide produces 1 mole of calcium carbonate

Since at NTP 1 mole of gas has a volume of 22.4 l or 22.4 dm3, it means 22.4 dm3 will produce 1 mole of calcium carbonate

Therefore 5.6 dm3 will produce 1X5.6/22.4=1/4 moles

Mass of 1 mole of CaCO3 = 40 + 12 +16 X3 = 100gms

so ¼ moles = 25 gms

Therefore 5.6 dm3 will produce 25 grams of CaCO3

Answered by  | 31st Jul, 2010, 04:36: PM

Ask Doubt

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.