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? 

Explanation

The reaction of sodium carbonate with acid is given as `Na_(2)CO_(3)+2HCl to 2NaCl+H_(2)O+CO_(2)`
1 mole of `Na_(2)CO_(3)="1 mole of "CO_(2)`
106 grams of `Na_(2)CO_(3)="44 grams of "CO_(2)`
5.3 grams of `Na_(2)CO_(3)=?`
The amount of `CO_(2)` produced `=(5.3)/(106) xx 44 =2.2"grams"`
Oxidation of graphite is given as `C+O_(2) to CO_(2)`
1 mole of `CO_(2)` =1 mole of graphite
44 grams of `CO_(2)=6.022 xx 10^(23)` atoms
2.2 grams of `CO_(2)=?`
Number of atoms of .C. to be oxidised `=(2.2 xx 6.022 xx 10^(23))/(44)=3.011 xx 10^(22)`