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5.3 grams of anhydrous sodium carbonate is treated with dilute hydrochloric acid to give carbondioxide. In order to produce the same amount of gas. How many graphite atoms are to be oxidised?

Answer» <html><body><p></p>Solution :The reaction of sodium <a href="https://interviewquestions.tuteehub.com/tag/carbonate-909380" style="font-weight:bold;" target="_blank" title="Click to know more about CARBONATE">CARBONATE</a> with acid is given as <br/> `Na_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)CO_(3)+2HClrarr 2NaCl+H_(2)O+CO_(2)` <br/> 1 <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of `Na_(2)CO_(3) = 1` mole of `CO_(2)` <br/> 106 grams of `Na_(2)CO_(3) = 44` grams of `CO_(2)` <br/> 5.3 grams of `Na_(2)CO_(3)=?` <br/> The <a href="https://interviewquestions.tuteehub.com/tag/amount-374803" style="font-weight:bold;" target="_blank" title="Click to know more about AMOUNT">AMOUNT</a> of `CO_(2)` produced `=(5.3)/(106)xx44=2.2` grams <br/> Oxidation of graphite is given as `C+O_(2)rarr CO_(2)` <br/> 1 mole of `CO_(2) = 1` mole of graphite <br/> 44 grams of `CO_(2) = 6.022 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(23)` atoms <br/> 2.2 grams of `CO_(2) = ?` <br/> Number of atoms of .C. to be oxidised <br/> `= (2.2xx6.022x10^(23))/(44)= 3.011xx10^(22)`</body></html>


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