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

Given conditions STP conditions
`P_(1)=750mm " "P_(2)=760mm`
`T_(1)=300K " "T_(2)=273K`
`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)=(750 xx 1 xx 273)/(300 xx 760)=0.989L`
Calcium cabonate on reaction with hydrochloric acid gives carbondioxide.
`CaCO_(3)+2HCl to CaCl_(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 `=(100 xx 0.898)/(22.4) =4.009 "grams"`