What weight of $BaCl_2$ would react with $24.4~\text{g}$ of sodium sulphate to produce $46.6~\text{g}$ of barium sulphate and $23.4~\text{g}$ of sodium chloride?
(Given, $BaCl_2+Na_2SO_4\to BaSO_4+2NaCl$)

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Solution

Let, the mass of $BaCl_2$ be $x~\text{g}$
According to the equation,
$\underset{x~\text{g}}{BaCl_2}+\underset{24.4~\text{g}}{Na_2SO_4}\to\underset{46.6~\text{g}}{ BaSO_4}+\underset{23.4~\text{g}}{2NaCl}$
According to law of conservation of mass,
$\text{Total mass of reactants=Total mass of products}...(i)$
$\text{Total mass of reactants=(x+24.4) g}$
$\text{Total mass of products=(46.6+23.4) g}$
Substituting in equation $(i)$
$x+24.4=46.6+23.4\\
x=46.6+23.4-24.4\\
\Rightarrow x=45.6~\text{g}$
Hence, the weight of $BaCl_2$ is $45.6~\text{g}$.

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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. The weight of BaCl2 that would react with 24.4g of sodium sulphate to produce 46.6g of barium sulphate and 23.4g of sodium chloride is ?

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 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 given below:
$N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 N a C l$
Calculate the percentage of sodium sulphate in the original mixture.

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What weight of \(BaCl_2\) would react with \(24.4~\text{g}\) of sodium sulphate to produce \(46.6~\text{g}\) of barium sulphate and \(23.4~\text{g}\) of sodium chloride? (Given, \(BaCl_2+Na_2SO_4\to BaSO_4+2NaCl\))

What weight of \(BaCl_2\) would react with \(24.4~\text{g}\) of sodium sulphate to produce \(46.6~\text{g}\) of barium sulphate and \(23.4~\text{g}\) of sodium chloride?
(Given, \(BaCl_2+Na_2SO_4\to BaSO_4+2NaCl\))