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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» <html><body><p></p>Solution :The corresponding chemical <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> is: <br/> `underset(40.08 +12.01 + 48.0=100.1)(CaCO_(3)) + underset(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> xx 36.46= 72.92)(2HCl) to CaCl_(2) + H_(2)O + underset("22.4 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a> at S.T.P")(CO_(2))` <br/> The mass of HCl present in `50.0 cm^(3)` of `M/2` of HCl <br/> `=(1/2 xx 36.46 xx 50.0)/1000 = 0.911 g` <br/> From the equation it is clear that 1 g of `CaCO_3` will require 0.73 g of <a href="https://interviewquestions.tuteehub.com/tag/hci-479434" style="font-weight:bold;" target="_blank" title="Click to know more about HCI">HCI</a>. Therefore, the mass of HCI used is more than the required mass. This implies that `CaCO_3` is the limiting reagent. <br/> `therefore 100.1 g` of `CaCO_(3)` give a volume of `CO_(2) = 22.4 L` <br/> `therefore 1.0 g` of `CaCO_(3)` give a volume of `CO_(2) = 22.4 L` <br/> `therefore 1.0 g` of `CaCO_(3)` will give a volume of `CO_(2)` <br/> `=22.4/100.1 xx 1.0 = 0.22 L` <br/> Hence,the volumeof carbondioxide evolved =0.22 L.</body></html>


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