This question is about redox reactions.
Sodium chlorate(V), NaClO3\text{NaClO}_3NaClO3, is an ionic compound containing the chlorate(V) ion, ClO3−\text{ClO}_3^-ClO3−. It can be prepared by reacting chlorine gas with hot concentrated sodium hydroxide solution:
3Cl2(g)+6NaOH(aq)→NaClO3(aq)+5NaCl(aq)+3H2O(l) 3\text{Cl}_2(\text{g}) + 6\text{NaOH}(\text{aq}) \rightarrow \text{NaClO}_3(\text{aq}) + 5\text{NaCl}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) 3Cl2(g)+6NaOH(aq)→NaClO3(aq)+5NaCl(aq)+3H2O(l)864 dm3864\text{ dm}^3864 dm3 of chlorine gas, measured at RTP, is reacted with an excess of hot concentrated sodium hydroxide solution. The resulting solid products are dissolved in water to form 1.60 m31.60\text{ m}^31.60 m3 of solution.
Calculate the concentration of NaClO3(aq)\text{NaClO}_3(\text{aq})NaClO3(aq) in this solution, in mol dm−3\text{mol dm}^{-3}mol dm−3.
Give your answer to an appropriate number of significant figures and in standard form. (Molar volume of gas at RTP = 24.0 dm3 mol−124.0\text{ dm}^3\text{ mol}^{-1}24.0 dm3 mol−1)
When solid sodium chlorate(III), NaClO2\text{NaClO}_2NaClO2, is heated, it undergoes a disproportionation reaction to form sodium chloride, NaCl\text{NaCl}NaCl, and sodium chlorate(V), NaClO3\text{NaClO}_3NaClO3.
Write an equation for this decomposition and explain, using oxidation numbers, why this is a disproportionation reaction.
