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.)

Answer 1

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