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 = $\frac{Mass}{Molar Mass}$
So,

No. of moles of MgCO₃ = $\frac{8.4 gm}{84.3 gm/mol}$ = 0.099 or 0.1 mol

​Therefore, 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