Question

$20.0 \,g$ 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 ?

• A. 60
• B. 84
• C. 75
• D. 96

Answer 1

Answer: (B)

$MgCO_{3} (s) \rightarrow MgO (s) + CO_{2} (g)$
moles of $MgCO_{3} = \frac{20}{84} = 0.238 $ mol
From above equation
1 mole $MgCO_3$ gives 1 mole MgO
$\therefore 0.238$ mole $MgCO_3$ will give $0.238$ mole $MgO$
$= 0.238 \times 40\, g = 9.523\, g\, MgO$
Practical yield of $MgO = 8 \,g\, MgO$
$\therefore $ % purity = $\frac{8}{9.523} \times 100 = 84\%$