The reactions of chlorine — practice questions

1. Write the balanced equation: $Cl_{2}(aq) + 2NaOH(aq) \to NaCl(aq) + NaClO(aq) + H_{2}O(l)$
2. Identify the oxidation state of chlorine in the reactant. In $Cl_{2}$, the oxidation state of each chlorine atom is 0.
3. Identify the oxidation states of chlorine in the products.
   • In sodium chloride (NaCl), the oxidation state of chlorine is -1.
   • In sodium chlorate(I) (NaClO), let the oxidation state of Cl be $x$. Na is +1 and O is -2. So, $(+1) + x + (-2) = 0$, which gives $x = +1$.
4. Conclude by comparing the states. The oxidation state of chlorine has decreased from 0 to -1 (reduction) and increased from 0 to +1 (oxidation). Since chlorine is simultaneously oxidised and reduced in the same reaction, it has undergone disproportionation.

1. Write the balanced equation: $Cl_{2}(g) + 2NaOH(aq) \to NaCl(aq) + NaClO(aq) + H_{2}O(l)$
2. Calculate the moles of chlorine gas: Moles = Volume / Molar Volume Moles of $Cl₂ = 1.20 , dm³ / 24.0 , dm³ , mol⁻¹ = 0.0500 , mol$
3. Use the molar ratio from the equation to find the moles of sodium chlorate(I): The ratio of $Cl_{2} : NaClO$ is 1 : 1. Therefore, moles of $NaClO = 0.0500 , mol$.
4. Calculate the molar mass of sodium chlorate(I), NaClO: $M_r(NaClO) = 23.0 + 35.5 + 16.0 = 74.5 , g , mol⁻¹$
5. Calculate the mass of sodium chlorate(I) produced: Mass = Moles × Molar Mass Mass of $NaClO = 0.0500 , mol × 74.5 , g , mol⁻¹ = 3.725 , g$

Answer: 3.73 g (to 3 significant figures).