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Question

20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample ?
(At. Wt. : Mg = 24)

A
60
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B
84
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C
75
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D
96
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Solution

The correct option is B 84
MgCO3(s)MgO(s)+CO2(g)
Moles of MgCO3=2084 mol
From the above equation
1 mole of MgCO3 gives 1 mole MgO
2084×40 g=20021 g MgO
Practical yield of MgO=8 gMgO
% purity =8×21200×100=84%

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