What is the weight of calcium carbonate required for the production of 1 L of carbon dioxide at `27^(@)C` and 750 mm, by the action of dilute hydrochloric acid?

Explanation

Givne conditions STP conditions
`{:(P_(1)="750 mm",P_(2)="760 mm"),(T_(1)="300 K",T_(2)="273 K"),(V_(1)="1 Lit",V_(2)=?):}`
From the equation of state, the unknown volume is obtained
`V_(2)=(P_(1)V_(1))/(T_(1))xx(T_(2))/(P_(2))=(750xx1xx273)/(300xx760)=0.898L`
Calcium carbonate on reaction with hydrochloric acid gives carbondioxide.
`CaCO_(3)+2HClrarrCaCl_(2)+H_(2)O+CO_(2)`
`"1 mole of "CO_(2)-="1 mole of "CaCO_(3)`
`"22.4 L of "CO_(2)"at STP"="100 grams of "CaCO_(3)`
`"0.898 L of "CO_(2)" at STP = ?"`
The weight of calcium carbonate required =
`(100xx0.989)/(*22.4)="4.009 grams"`