One mole of magnesium carbonate is subjected to themal decomposition in an open vessel. Percentage loss in weight suffered will be

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46.8%

61.2%

44%

52.4%

answer is C.

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Detailed Solution

$\begin{array}{l} {{MgCO}_{3} (s)}_{84 g}^{1 mole} \to MgCO (s) + {{CO}_{2} (g)}_{44 g}^{1 mole} \\ {CO}_{2} (g) escapes out . \\ weight loss = 44 g \\ % loss in weight = \frac{44}{84} \times 100 = 52 .4 % \end{array}$

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