1g of a mixture containing `Na_(2)CO_(3) and NaHCO_(3)` in equal amounts is heated until a constant mass is obtained. Find the volume of `CO_(2)` gas liberated at STP.

Explanation

Amount of `NaHCO_(3)` in the mixture=0.5
On heating `NaHCO_(3)` dissociates to give `Na_(2)CO_(3),CO_(2) and H_(2)O,Na_(2)CO_(3) ` dissociates to give `Na_(2)CO_(3)`,
`underset(2xx85g)(2NaHCO_(3)(s)) to Na_(2)CO_(3)(s)+underset(22400mL[STP])(CO_(2)(g)+H_(2)O)`
`therefore `168g `NaHCO_(3)-=22400mL " "CO_(2)(STP)`
`therefore 0.5" "NaHCO_(3)-=66.67mL," So, "CO_(2)` liberated at STP= 66.67 mL.