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)rarrMgO+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)xx84=4.2` grams
6 grams mineral has 4.2 grams `MgCO_(3)`
 Percent purity `=(4.2)/(6)xx100=70%`