20.0 g of a magnesium carbonate sample decomposes on heating to give c

1.	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» Correct option: (B) 84 MgCO_3(s) → MgO_(s) + CO_2(g) Molar mass of MgCO₃ = 84 g mol^-1 ∴ Number of moles of MgCO₃ = \(\cfrac{20}{84}\) = 0.238 mol ∵ 1 mole MgCO₃ gives 1 mole MgO ∴ 0.238 mole MgCO₃ will give 0.238 mole MgO. Molar mass of MgO = 40 g mol^-1 ∴ 0.238 mole MgO = 40 x 0.238 = 9.52 g MgO ∴ Theoretical yield of MgO = 9.52 g Practical yield of MgO is 8.0 g ∴ Percentage purity = \(\cfrac{8}{9.52}\) x 100 = 84%

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»

Correct option: (B) 84

MgCO_3(s) → MgO_(s) + CO_2(g)

Molar mass of MgCO₃ = 84 g mol^-1

∴ Number of moles of MgCO₃ = \(\cfrac{20}{84}\)

= 0.238 mol

∵ 1 mole MgCO₃ gives 1 mole MgO

∴ 0.238 mole MgCO₃ will give 0.238 mole MgO.

Molar mass of MgO = 40 g mol^-1

∴ 0.238 mole MgO = 40 x 0.238

= 9.52 g MgO

∴ Theoretical yield of MgO = 9.52 g

Practical yield of MgO is 8.0 g

∴ Percentage purity = \(\cfrac{8}{9.52}\) x 100

= 84%