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PROBLEM: Aqueous Sodium Hydroxide plus aqueous Calcium Hydroxide plus solid Carbon plus Chlorine dioxide gas yields aqueous Sodium Chlorite plus solid.

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Presentation on theme: "PROBLEM: Aqueous Sodium Hydroxide plus aqueous Calcium Hydroxide plus solid Carbon plus Chlorine dioxide gas yields aqueous Sodium Chlorite plus solid."— Presentation transcript:

1 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) 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. 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. Step 3: Write and balance the oxidation half reaction. Note – since CaCO 3 is a solid it does NOT dissociate C --------> CaCO 3 3H 2 O + Ca 2+ + C --------> CaCO 3 Since the C is already balanced, balance the Ca by adding Ca 2+ Ca 2+ + C --------> CaCO 3 Next add 3 H 2 O to balance the O mass 3H 2 O + Ca 2+ + C --------> CaCO 3 + 6 H + + 4 e – Next add e – to balance the charge – since the left is +2, and the right is +6, add 4e– to the right NaOH(aq) + Ca(OH) 2 (aq) + C(s) + ClO 2 (g) --------> NaClO 2 (aq) + CaCO 3 (s) + H 2 O(l) +1 +2–2+3–20+4+1 –2 +2+4–2+1–2 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 3H 2 O + Ca 2+ + C --------> CaCO 3 + 6 H + Next balance the H mass by adding 6H +

2 Step 5: Next make the oxidation and reduction half-reactions have the same # of e – ClO 2 (g) --------> ClO 2 – Step 4: Write and balance the reduction half reaction Mass is already balanced so add e – to balance the charge – just add 1 e – e – + ClO 2 (g) --------> ClO 2 – 3H 2 O + Ca 2+ + C --------> CaCO 3 + 6 H + + 4 e – e – + ClO 2 --------> ClO 2 – (X4) 4 e – + 4 ClO 2 --------> 4 ClO 2 – 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 H 2 O + Ca 2+ + C + 4 ClO 2 --------> 4 ClO 2 – + CaCO 3 + 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 H 2 O + Ca 2+ + C + 4 ClO 2 --------> 4 ClO 2 – + CaCO 3 + 6 H + + 6 OH – Step 8: OH – plus H + makes water – combine and cancel out waters from both sides 6 OH – + 3 H 2 O + Ca 2+ + C + 4 ClO 2 --------> 4 ClO 2 – + CaCO 3 + 6 H 2 O 6 OH – + Ca 2+ + C + 4 ClO 2 --------> 4 ClO 2 – + CaCO 3 + 3 H 2 O

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


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