Question

$20.0g$ of a magnesium carbonate sample decomposes on heating to give carbon dioxide and $8.0g$ 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)

$MgC{O}_3(s)\to MgO(s)+C{O}_2(g)$
moles of $MgC{O}_3=\frac{20}{84}=0.238$ mol
From above equation
1 mole $MgC{O}_3$ gives 1 mole MgO
$\therefore 0.238$ mole $MgC{O}_3$ will give $0.238$ mole $MgO$
$=0.238\times 40g=9.523gMgO$
Practical yield of $MgO=8gMgO$
$\therefore$ % purity = $\frac{8}{9.523}\times 100=84\%$