A mixture of sodium chloride and sodium iodide when treated with sulphuric acid give sodium sulphate equal to the weight of original mixture. Find out percentage composition of sodium chloride and sodium iodide.

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Write the balanced chemical equations for the following reactions-

(a) Sodium carbonate on reaction with hydrochloric acid in equal molar concentration gives sodium chloride and sodium hydrogen carbonate.

(b) Sodium hydrogen carbonate on reaction with hydrochloric acid gives sodium chloride, water and liberates carbon dioxide.

(c) Copper sulphate on treatment with potassium iodide precipitates cuprous iodide $({Cu}_{2} I_{2})$ , liberates iodine gas and also forms potassium sulphate.

Q. 10 g of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g. of barium sulphate is precipitated according to the equation :-

N a_{2} S O_{4} + B a C l_{2} ⟶ B a S O_{4} + 2 N a C l

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The percentage of sodium sulphate in the original mixture is:

[N a = 23, O = 16, S = 32, B a = 132]

Q. 3.6 g mixture of sodium chloride and potassium chloride is dissolved in water. The solution is treated with excess of silver nitrate solution. 7.74 g of silver chloride is obtained. Find the percentage of sodium chloride and potassium chloride in the mixture.

Q. If 2 moles of NaOH reacts with 4 moles of HCl to give sodium chloride and then the produced sodium chloride reacts with 3 moles of concentrated sulphuric acid to give sodium sulphate, find the total moles of sodium sulphate produced. The concerned reactions are

N a O H + H C l (d i l u t e d) \to N a C l + H_{2} O

2 N a C l + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H C l

Q. How would you obtain sodium chloride from a mixture of sodium chloride and sulphur without using water ?

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