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Question

1.84g mixture of CaCO3and MgCO3​ was heated to a constant weight till 0.96g residue was formed.Percentage of MgCO3in sample was ?

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Solution

Number of moles of CaCO3 = XNumber of moles of MgCO3 = YMolar mass of CaCO3 = 100 gMolar mass of MgCO3 = 84.3 moleThus100 X + 84.3 Y = 1.84 ----- 1 (as the mixture weight is = 1.84 g)On heating the mixture CO2 is expelled out and both oxides remain in the residue.CaCO3 + MgCO3 CaO + MgO + 2 CO2Molar mass of CaO = 56 gMolar mass of MgO = 40.3 gAs the number of moles of oxides are same with carbonates56 X+ 40.3 Y = 0.96 ------ 2 (given tha the residual weight is 0.96 g)Solve equation 1 and 2We getX= 0.0098 and Y = 0.01Mass of CaCO3 = 100 × 0.0098 = 0.98 gMass of MgCO3 = 84.3 × 0.01 = 0.86 gMass % of CaCO3 = 0.981.84 × 100 = 53.26 %Mass % of MgCO3 = 0.861.84 × 100 = 46.74 %

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