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Step 1 : Write the unbalanced formula equations

NaOH ( aq ) + Ca (OH) 2 ( aq ) + C(s) + ClO 2 (g) --------> NaClO 2 ( aq ) + CaCO 3 (s) + H 2 O(l). Step 2 : Identify the species that are oxidized and reduced. Start by labeling the oxidation #s. +1. –2. +1. +2. –2. +1. 0. +4. –2. +1. +3. –2. +2. +4. –2. +1. –2.

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Step 1 : Write the unbalanced formula equations

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  1. NaOH(aq) + Ca(OH)2(aq) + C(s) + ClO2(g) --------> NaClO2(aq) + CaCO3(s) + H2O(l) Step 2: Identify the species that are oxidized and reduced. Start by labeling the oxidation #s. +1 –2 +1 +2 –2 +1 0 +4 –2 +1 +3 –2 +2 +4 –2 +1 –2 PROBLEM:Aqueous Sodium Hydroxide plus aqueous Calcium Hydroxide plus solid Carbon plus Chlorine dioxide gas yields aqueous Sodium Chlorite plus solid calcium carbonate plus liquid water. (Note: The reaction takes place in basic solution) NaOH(aq) + Ca(OH)2(aq) + C(s) + ClO2(g) --------> NaClO2(aq) + CaCO3(s) + H2O(l) Since C goes from 0 to +4 it lost e– and is oxidized. Since Cl goes from +4 to +3 it gained e– and is reduced. Now we will balance the equation as if it was in acid and then switch it to base after we have the net ionic equation Step 3: Write and balance the oxidation half reaction. Note – since CaCO3 is a solid it does NOT dissociate Step 1: Write the unbalanced formula equations C--------> CaCO3 Since the C is already balanced, balance the Ca by adding Ca2+ Ca2+ + C --------> CaCO3 Next add 3 H2O to balance the O mass 3H2O + Ca2++ C --------> CaCO3 Next balance the H mass by adding 6H+ 3H2O + Ca2++ C --------> CaCO3 + 6 H+ Next add e– to balance the charge – since the left is +2, and the right is +6, add 4e– to the right 3H2O + Ca2+ + C --------> CaCO3 + 6 H+ + 4 e–

  2. Step 4: Write and balance the reduction half reaction ClO2(g) --------> ClO2– Mass is already balanced so add e– to balance the charge – just add 1 e– e– + ClO2(g) --------> ClO2– Step 5: Next make the oxidation and reduction half-reactions have the same # of e– 3H2O + Ca2+ + C --------> CaCO3 + 6 H+ + 4 e– e– + ClO2--------> ClO2–(X4) 4 e– + 4 ClO2--------> 4 ClO2– Step 6: Combine the two half-reactions together, to make the net ionic equation. Remember to cancel out the items on opposite sides of arrow (water, H+ ions, and e–). In this case in only the e–cancel out. 3 H2O + Ca2+ + C + 4 ClO2 --------> 4 ClO2– + CaCO3 + 6 H+ Step 7: Convert to base by adding an OH– ion for every H+ ion. But remember – to keep the equation balanced you must add to both sides. 6 OH– + 3 H2O + Ca2+ + C + 4 ClO2 --------> 4 ClO2– + CaCO3 + 6 H+ + 6 OH– Step 8: OH– plus H+ makes water – combine and cancel out waters from both sides 6 OH– + 3 H2O + Ca2+ + C + 4 ClO2 --------> 4 ClO2– + CaCO3 + 6 H2O 6 OH– + Ca2+ + C + 4 ClO2 --------> 4 ClO2– + CaCO3 + 3 H2O

  3. Step 7: Add back spectator ions and combine with other ions to write complete compounds and the balanced overall equation. Combine Ca2+and OH–. Then add Na+ where needed to anions. 4 NaOH– + Ca(OH)2+ C + 4 ClO2 --------> 4 NaClO2+ CaCO3 + 3 H2O

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