10g of mixture sodium chloride and anhydrous sodium sulphate is dissolved in water.An excess of barium chloride solution is added and 6.99g of barium sulphate is precipitated according to the equation : Na2SO4 + BaCl2 -----> BaSO4 + 2NaCl. Calculate the percentage os sodium sulphate in the original mixture.

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Solution

1. Reaction involved :

$N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 N a C l$

2. Number of moles of barium sulphate obtained = $\frac{G i v e n m a s s}{M o l a r m a s s} = \frac{6.99 g}{233.43 g / m o l} = 0.0299 m o l e s$

3. Let mass of NaCl be x
and of Na₂SO₄ be 10-x

4. As 1 mole of Na₂SO₄ gives 1 mole of BaSO₄,

So, Number of moles of Na₂SO₄ = 0.0299moles

5. Mass of Na₂SO₄ present in mixture = $N o . o f m o l e s o f N a_{2} S O_{4} \times M o l a r m a s s = 0.0299 \times 142.04 = 4.24 g$

6. Percentage of sodium sulphate = $\frac{4.24}{10} \times 100 = 42.4 %$

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Similar questions

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

.
The percentage of sodium sulphate in the original mixture is:

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

10 g of a mixture of barium 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 given below:

Na₂SO₄ + BaCl₂ → BaSO₄ +2NaCl

[O = 16, Na = 23, S = 32, Ba = 137]

Calculate the percentage of sodium sulphate in the original mixture.

Barium chloride on reacting with sodium sulphate forms barium sulphate and sodium chloride. Which of the following is the precipitate?

Q. Barium chloride reacts with sodium sulphate to form Barium sulphate and sodium chloride. Then, according to the law of conservation of mass:

Q. What weight of

B a C l_{2}

would react with

24.4 g

of sodium sulphate to produce

46.6 g

of barium sulphate and

23.4 g

of sodium chloride?
(Given,

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

)

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