1.42 g of sodium sulphate dissolved in water was treated with excess of $B a C l_{2}$ solution and 1.167 g of $B a S O_{4}$ was obtained. Calculate the percentage yield of the reaction.
Molar mass of sodium sulphate and barium sulphate is 142 g/mol and 233.4 g/mole respectively.

100%

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50%

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75%

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90%

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Solution

The correct option is B 50%
Given
$N a_{2} S O_{4} + B a C l_{2} \to 2 N a C l + B a S O_{4}$
Mass of sodium sulphate = 1.42 g
Moles of sodium sulphate = $\frac{1.42}{142} = 0.01$
Theoretical yield of barium sulphate = 0.01
Mass of barium sulphate obtained = 1.167 g
Moles of barium sulphate = $\frac{1.167}{233.4} = 0.005$
Percentage yield = $\frac{experimental yield}{theoretical yield} \times 100 = \frac{0.005}{0.01} \times 100 = 50 %$

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Similar questions

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.

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]

Q. 11.65 g of

B a S O_{4}

was obtained as a dry precipitate when 10 g of an impure sample of

N a_{2} S O_{4}

was dissolved in water and treated with an excess of

B a C l_{2}

. Calculate the percentage purity of the

N a_{2} S O_{4}

sample. (Molar mass of

B a S O_{4} = 233 g a n d N a_{2} S O 4 = 142 g

)

Q. 1.42 g of sodium sulphate is dissolved in 100 mL water. 4.16 g of barium chloride is dissolved in another 100 mL water. The amount of precipitate of barium sulphate upon mixing of the two solutions is:

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

(Atomic mass: Na = 23, S = 32, O =16, Ba = 137, Cl = 35.5)

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

(Atomic mass: Na = 23, S = 32, O =16, Ba = 137, Cl = 35.5)

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1.42 g of sodium sulphate dissolved in water was treated with excess of BaCl2 solution and 1.167 g of BaSO4 was obtained. Calculate the percentage yield of the reaction. Molar mass of sodium sulphate and barium sulphate is 142 g/mol and 233.4 g/mole respectively.

1.42 g of sodium sulphate dissolved in water was treated with excess of $B a C l_{2}$ solution and 1.167 g of $B a S O_{4}$ was obtained. Calculate the percentage yield of the reaction.
Molar mass of sodium sulphate and barium sulphate is 142 g/mol and 233.4 g/mole respectively.