MgCO3(s)⟶MgO(s)+CO2(g)
MolesofMgCO3=8420=0.238mol
From above equation.
1 mole MgCO3 gives 1 mole MgO
∴ 0.238 mole MgCO3 will give 0.238 mole MgO
=0.238×40g=9.523gMgO
PracticalyieldofMgO=8gMgO
∴
NEET UG 2015 — Chemistry Physical Chemistry
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)
Held on 30 Apr 2015 · Verified 9 Jul 2026.
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