20.0 g of a magnesium carbonate sample decom | vedclass

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(A) The decomposition reaction is: $MgCO_{3(s)} \longrightarrow MgO_{(s)} + CO_{2(g)}$The molar mass of $MgO = 24 + 16 = 40 \ g/mol$.The number of mol

Vedclass · Accepted Answer

$63$

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(A) The decomposition reaction is: $MgCO_{3(s)} \longrightarrow MgO_{(s)} + CO_{2(g)}$The molar mass of $MgO = 24 + 16 = 40 \ g/mol$.The number of moles of $MgO$ produced $= \frac{6.0 \ g}{40 \ g/mol} = 0.15 \ mol$.According to the stoichiometry,$1 \ mol$ of $MgCO_3$ produces $1 \ mol$ of $MgO$.Therefore,the moles of pure $MgCO_3$ present $= 0.15 \ mol$.The molar mass of $MgCO_3 = 24 + 12 + (3 	imes 16) = 84 \ g/mol$.The mass of pure $MgCO_3 = 0.15 \ mol 	imes 84 \ g/mol = 12.6 \ g$.The percentage purity $= \frac{	ext{mass of pure } MgCO_3}{	ext{total mass of sample}} 	imes 100 = \frac{12.6}{20.0} 	imes 100 = 63 \%$.

$20.0 \ g$ of a magnesium carbonate sample decomposes on heating to give carbon dioxide and $6.0 \ g$ magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (Atomic weight of $Mg = 24$)

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