8.4 g MgCO3 on heating leaves behind a residue weighing 4g carbon dioxide released into atmosphere at STP will be

Dear Student


Mass of MgCO3 = 8.4 g

Mass of Residue = 4 g

The reaction involved is -

MgCO3 ---> MgO + CO2

No. of moles = MassMolar Mass
So,

No. of moles of MgCO38.4 gm84.3 gm/mol = 0.099 or 0.1 mol

‚ÄčTherefore, Mass of CO2 = No. of moles x molar mass

                                       = 0.1 mol x 44 gm/mol = 4.4 gm


Another method-

According to the law of conservation of mass-

Mass of reactants = Mass of products

Mass of MgCO3 = Mass of (MgO + CO2)

8.4 g = 4 gm - mass of CO2

So, Mass of CO2 = 8.4 - 4 = 4.4 gm



Regards

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