100 g of a pure sample of CaCO_3 is treated a with 500 cm^(3) of M/2HCI solution. Calculate the of CO_2 that will be evolved at S.T.P.

100 g of a pure sample of CaCO_3 is treated a with 500 cm^(3) of M/2HC

1.	100 g of a pure sample of CaCO_3 is treated a with 500 cm^(3) of M/2HCI solution. Calculate the of CO_2 that will be evolved at S.T.P.
Answer» Solution :The corresponding chemical EQUATION is: `underset(40.08 +12.01 + 48.0=100.1)(CaCO_(3)) + underset(2 xx 36.46= 72.92)(2HCl) to CaCl_(2) + H_(2)O + underset("22.4 L at S.T.P")(CO_(2))` The mass of HCl present in `50.0 cm^(3)` of `M/2` of HCl `=(1/2 xx 36.46 xx 50.0)/1000 = 0.911 g` From the equation it is clear that 1 g of `CaCO_3` will require 0.73 g of HCI. Therefore, the mass of HCI used is more than the required mass. This implies that `CaCO_3` is the limiting reagent. `therefore 100.1 g` of `CaCO_(3)` give a volume of `CO_(2) = 22.4 L` `therefore 1.0 g` of `CaCO_(3)` give a volume of `CO_(2) = 22.4 L` `therefore 1.0 g` of `CaCO_(3)` will give a volume of `CO_(2)` `=22.4/100.1 xx 1.0 = 0.22 L` Hence,the volumeof carbondioxide evolved =0.22 L.