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Basic Buffer Action

The amount of...

Question

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)

23.3

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11.65

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30.6

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33.2

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Solution

The correct option is C $11.65$ g
$100$ ml of $20.8 %$ $B a C l_{2}$ solution $= 20.8$ g $B a C l_{2} =$ $0.1$ mol.
$50$ ml of $9.8 %$ $H_{2} S O_{4}$ solution $= 4.9$ g $H_{2} S O_{4} =$ $0.05$ mol.
The reaction is as follows:
$B a C l_{2} + H_{2} S O_{4} ⇋ B a S O_{4} + 2 H C l$
Since, $H_{2} S O_{4}$ is the limiting reagent, only $0.05$ moles of $B a S O_{4}$ will form.
Hence, $0.05 \times 233 = 11.65$ g of $B a S O_{4}$ is formed.

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Similar questions

B a S O_{4}

100

20.8 %

B a C l_{2}

50

9.8 %

H_{2} S O_{4}

= 137

= 35.5

= 32

= 1

= 16

B a C l_{2}

H_{2} S O_{4}

B a S O_{4}

(B a = 137, C l = 35.5, S = 32, H = 1, O = 16)

B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l

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

B a C l_{2} (a q)

H_{2} S O_{4}

B a S O_{4}

500

0.25 M N a_{2} S O_{4}

15.0

B a C l_{2}

B a S O_{4}

B a S O_{4}

y \times 10^{- 3}

y

B a = 137, S = 32, O = 16, N a = 23

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