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?

5.3 grams of anhydrous sodium carbonate is treated with dilute hydroch

1.	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» Solution :The reaction of sodium CARBONATE with acid is given as `Na_(2)CO_(3)+2HClrarr 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)xx44=2.2` grams Oxidation of graphite is given as `C+O_(2)rarr 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 `= (2.2xx6.022x10^(23))/(44)= 3.011xx10^(22)`