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

25.2 g

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11.2 g

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45.6 g

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50.0 g

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Solution

The correct option is C $45.6 g$
Let, the mass of $B a C l_{2}$ be $x g$
According to the equation,
$B a C l_{2} x g + N a_{2} S O_{4} 24.4 g \to B a S O_{4} 46.6 g + 2 N a C l 23.4 g$
According to law of conservation of mass,
$Total mass of reactants=Total mass of products . . . (i)$
$Total mass of reactants=(x+24.4) g$
$Total mass of products=(46.6+23.4) g$
Substituting in equation $(i)$
$x + 24.4 = 46.6 + 23.4 x = 46.6 + 23.4 - 24.4 \Rightarrow x = 45.6 g$
Hence, the weight of $B a C l_{2}$ is $45.6 g$ .

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Similar questions

Q. What weight of

B a C l_{2}

(in grams) 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

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. 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)

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. 142 g of sodium sulphate

(N a_{2} S O_{4})

has:

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What weight of BaCl2 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, BaCl2+Na2SO4→BaSO4+2NaCl)

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