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Limiting Reagent or Reactant

1.42 g of sod...

Question

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)

2.33 g

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4.66 g

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2.52 g

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1.42 g

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Solution

The correct option is A 2.33 g
Molar mass of sodium sulphate $(N a_{2} S O_{4}) = 23 \times 2 + 32 + 16 \times 4 = 142 g / m o l$

Number of moles of sodium sulphate = $\frac{1.42}{142} = 0.01 m o l$

Molar mass of barium chloride $(B a C l_{2}) = 137 + 35.5 \times 2 = 137 + 71 = 208 g / m o l$

Number of moles of barium chloride = $\frac{4.16}{208} = 0.02 m o l$

Hence, sodium sulphate is the limiting reagent.

1 mole of sodium sulphate gives 1 mole of barium sulphate.

Hence, 0.01 mole of sodium sulphate gives 0.01 mole of barium sulphate.

Molar mass of barium sulphate $(B a S O_{4}) = 137 + 32 + 16 \times 4 = 233 g / m o l$

Hence, mass of the barium sulphate precipitate formed = $0.01 \times 233 = 2.33 g$

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Similar questions

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)

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.

Q. The amount of

B a S O_{4}

formed upon mixing

100

mL of

20.8 %

B a C l_{2}

solution with

50

mL of

9.8 %

H_{2} S O_{4}

solution will be:
[Ba

= 137

, Cl

= 35.5

, S

= 32

, H

= 1

and O

= 16

Q. The amount of

B a S O_{4}

formed upon mixing

100

ml of

20.8 %

B a C l_{2}

solution with

50

ml of

9.8 %

H_{2} S O_{4}

solution will be:

(Given that molecular weight of

B a = 137, C l = 35.5, S = 32, H = 1

and

O = 16

g/mol)

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Limiting Reagent or Reactant

Standard XII Chemistry

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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: Na2SO4+BaCl2→BaSO4+2NaCl (Atomic mass: Na = 23, S = 32, O =16, Ba = 137, Cl = 35.5)