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)

84

Answer 2

${MgCO_{3} (s) -> MgO (s) + CO_{2} (g)}$
moles of ${MgCO_{3} = \frac{20}{84} = 0.238 } \text{mol}$
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%