Question

20 g of magnesium carbonate decomposes on heating to give carbon dioxide n 8g of magnesium oxide wglhat will ne the percentage purity f magnesium carbonate in this sample??

Answer 1

The equation representing decomposition of Magnesium carbonate:

$MgCO_{3} (s) ---> MgO (s) + CO_{2}(g)$

Moles of MgO = $8 g MgO * \frac{1 mol MgO}{40.30 g MgO} </p><p> =0.199 mol MgO$

Moles of $MgCO_{3}$ = $0.199 mol MgO * \frac{1 mol MgCO_{3}}{1 mol MgO}$

= 0.199 mol $MgCO_{3}$

Mass of pure $MgCO_{3}$ = $0.199 mol MgCO_{3} * \frac{84.31 g MgCO_{3}}{1 mol MgCO_{3}}$ = 16.74 g

% purity of Mg$CO_{3}$ in the sample =$\frac{16.74 g}{20 g}*100$ = 84 %