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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.: Mg = 24)
A
60
B
84
C
75
D
96
Detailed Solution
$MgCO_3(s) \rightarrow MgO(s) + CO_2(g)$
Moles of $MgCO_3$ if pure = $\frac{20}{84} = 0.238$ mol, which would give $0.238\times 40 = 9.523$ g MgO.
Actual MgO = 8 g
% purity = $\frac{8}{9.523}\times 100 = 84\%$
Moles of $MgCO_3$ if pure = $\frac{20}{84} = 0.238$ mol, which would give $0.238\times 40 = 9.523$ g MgO.
Actual MgO = 8 g
% purity = $\frac{8}{9.523}\times 100 = 84\%$