Question

Calculate the amount of 80% pure NaOH sample required to react completely with 21.3 g of Chlorine in hot condition?

Answer 1

30 g

Answer 2

The reaction of chlorine with hot, concentrated sodium hydroxide is a disproportionation reaction. The balanced chemical equation is: $3 Cl_2(g) + 6 NaOH(aq) \xrightarrow{hot, conc.} 5 NaCl(aq) + NaClO_3(aq) + 3 H_2O(l)$

1. Moles of Chlorine ($Cl_2$): Mass of $Cl_2 = 21.3$ g. Molar mass of $Cl_2 = 71$ g/mol. Moles of $Cl_2 = \frac{21.3 \text{ g}}{71 \text{ g/mol}} = 0.3$ mol.
2. Moles of NaOH required: From the balanced equation, the mole ratio of $Cl_2$ to NaOH is $3:6$ ($1:2$). Moles of NaOH required = $2 \times (\text{moles of } Cl_2) = 2 \times 0.3$ mol $= 0.6$ mol.
3. Mass of pure NaOH required: Molar mass of NaOH = $40$ g/mol. Mass of pure NaOH required = $0.6 \text{ mol} \times 40 \text{ g/mol} = 24$ g.
4. Mass of 80% pure NaOH sample: Let the required mass of the 80% pure NaOH sample be $m_{sample}$. $0.80 \times m_{sample} = 24$ g $m_{sample} = \frac{24 \text{ g}}{0.80} = 30$ g.