6 grams of magnesite mineral on heating liberated carbondioxide which measures 1.12 L at STP. What is the percentage purity of mineral?

6 grams of magnesite mineral on heating liberated carbondioxide which

1.	6 grams of magnesite mineral on heating liberated carbondioxide which measures 1.12 L at STP. What is the percentage purity of mineral?
Answer» Solution :CHEMICAL constituent of magnesite is magnesium carbonate. On heating it LIBERATES carbondioxide. `MgCO_(3) to MgO+CO_(2)` 1 mole of `CO_(2)="1 mole of "MgCO_(3)` 22.4 L of `CO_(2)` at STP=84 grams of `MgCO_(3)` 1.12 L of `CO_(2)` at STP=? The WEIGHT of pure `MgCO_(3)` to be decomposed `(1.12)/(22.4) xx 84 =4.2"grams"` Percent purity `=(4.2)/(6) xx 100=70%`