Magnesium carbonate (MgCO3) and calcium carbonate (CaCO3) both decompose when he

Question

Magnesium carbonate (MgCO3) and calcium carbonate (CaCO3) both decompose when heated, forming CO2 gas and the corresponding oxides.

MgCO3(s) CO2(g) + MgO(s)

CaCO3(s) CO2(g) + CaO(s)

When a particular mixture of MgCO3 and CaCO3 is heated, it releases 47% of its mass as CO2, so that the oxide products have 53%of the mass of the original sample. What mass percentage of MgCO3 was present in the original mixture? (HINT: The fact that the sample loses 47% of its mass does not depend on the original sample size, only on the relative amounts of the two carbonates. So you can start with any size sample you’d like.)

Explanation / Answer

now

let the mass of sample be 1000 g

given

oxides weigh 53 % of the original sample

so

mass of oxides = 530 g

that is

mass of MgO + mass of Ca0 = 530 g

now

given

47% of the mass is emitted as C02

so

mass of C02 emitted = 470 g

now

moles of C02 emitted = mass / molar mass

moles of C02 emitted = 470 / 44

moles of C02 emitted = 10.6818

now

let

y be the moles of MgC03

then

moles of C02 emitted from MgC03 = y

moles of C02 emitted from CaC02 = 10.6818 - y

so

moles of CaC03 = 10.6818 - y

moles of Mg0   = y

mass of MgO = y x 40

mass of Mg0 = 40 y

so

mass of CaO = 530 - 40 y

moles of CaO = ( 530 -40y) / 56

so

moles of CaC03 = ( 530-40y) / 56

so

(530-40y) / 56 = 10.6818 - y

530 -40y = 598.1818 - 56y

y = 4.26

so

moles of CaC03 = 4.26

mass of CaC03 = moles x molar mass

mass of CaC03 = 4.26 x 100

mass of CaC03 = 426

so

mass of MgC03 = 1000 - 426

mass of MgC03 = 574

so

% MgC03 = 574 x 100 / 1000

% mgC03 = 57.4

so

mass percentagge of MgC03 is 57.4

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