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

Explanation

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%`