What amount of `CaCO_(3)` will be just sufficient to produce 1 L of `CO_(2)` at `27^(@)C` and 760 mmHg pressure? Determine the quantity of pure carbon requried to yield the same amount of `CO_(2)`.

Explanation

Let, at `27^(@)C` and 760 mm Hg pressure volume of 1L `CO_(2)` gas at STP be V L, then
`(Vxx760)/(273)=(1xx760)/((273+27))[(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))]`
`therefore V=(273)/(300)=0.91L`
Mass of 0.91L of `CO_(2)` gas at `STP=(44)/(22.4)xx0.91=1.7875g`
`underset(100g)(CaCO_(3))(s) to CaO(s)+ underset(44g)(CO_(2)(g))`
`therefore` 44g `CO_(2)-=100g" "CaCO_(3)` or,
`1.7875g" "CO_(2)-=4.0625g" "CaCO_(3)`
`therefore` At `27^(@)C` and 760 mm Hg pressure 4.0625 `CaCO_(3)` is requried to produce 1L `CO_(2)` gas at STP.
`underset(12g)(C(s))+O_(2)(g) to underset(44g)(CO_(2)(g))`
`therefoe` 44g `CO_(2)-=12g" "C" or,"1.7875g" "CO_(2)-=0.4875g` C.