20.0 g of a magnesium carbonate sample decom | vedclass

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(C) The decomposition reaction is: $MgCO_{3(s)} \longrightarrow MgO_{(s)} + CO_{2(g)}$Molar mass of $MgCO_3 = 24 + 12 + (3 	imes 16) = 84 \ g/mol$.Mo

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$84$

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(C) The decomposition reaction is: $MgCO_{3(s)} \longrightarrow MgO_{(s)} + CO_{2(g)}$Molar mass of $MgCO_3 = 24 + 12 + (3 	imes 16) = 84 \ g/mol$.Molar mass of $MgO = 24 + 16 = 40 \ g/mol$.According to the stoichiometry,$84 \ g$ of pure $MgCO_3$ produces $40 \ g$ of $MgO$.Therefore,the mass of pure $MgCO_3$ required to produce $8.0 \ g$ of $MgO$ is: $\frac{84 \ g \ MgCO_3}{40 \ g \ MgO} 	imes 8.0 \ g \ MgO = 16.8 \ g \ MgCO_3$.Percentage purity of the sample = $\frac{	ext{Mass of pure } MgCO_3}{	ext{Total mass of sample}} 	imes 100 = \frac{16.8}{20.0} 	imes 100 = 84 \ \%$.

$20.0 \ g$ 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? $(At. \ wt. \ of \ Mg = 24)$

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