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Concentration Terms - An Introduction (w/w etc)

500 mL of 0...

Question

$500$ mL of $0.25 M N a_{2} S O_{4}$ solution is added to an aqueous solution of $15.0$ g of $B a C l_{2}$ solution resulting in the formation of white precipitate of $B a S O_{4}$ . The moles of $B a S O_{4}$ formed is $y \times 10^{- 3}$ moles. Then, $y$ is (in nearest integer) : [Given that molecular weight of $B a = 137, S = 32, O = 16, N a = 23$ g/mol]

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Solution

The reaction is as follows:
$B a C l_{2} + N a_{2} S O_{4} \to B a S O_{4} + 2 N a C l$
The number of moles of sodium sulfate $= 0.5 \times 0.25 = 0.125$ mol
The mass of barium chloride is $15$ g. This corresponds to $0.0721$ moles. Hence, barium chloride is the limiting reactant.
The number of moles of barium sulphate formed is equal to the number of moles of barium chloride.
It is equal to $0.0721$ mol $= 72 \times 10^{- 3}$
Thus, the value of y is 72.

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