20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium…

20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample?


(Atomic weight: Mg = 24)
  1. 75
  2. 96
  3. 60
  4. 84

Solution

MgCO3sMgOs+CO2g
Moles of MgCO3=2084=0.238 mol
From above equation.
1 mole MgCO3 gives 1 mole MgO
0.238 mole MgCO3 will give 0.238 mole MgO
=0.238×40 g=9.523 g MgO
Practical yield of MgO=8 g MgO
% Purity=89.523×100=84%

Asked in: MHT CET Full Test 2

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