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Limiting Reagent or Reactant

10 mL of 1 ...

Question

$10$ mL of $1 M B a C l_{2}$ solution and $5$ mL of $0.5 M K_{2} S O_{4}$ are mixed together to percipitate out $B a S O_{4}$ . The amount of $B a S O_{4}$ precipated will be :

0.005

mol

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0.00025

mol

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0.025

mol

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0.0025

mol

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Solution

The correct option is C $0.0025$ mol
The reaction is as follows:

$B a C l_{2} + K_{2} S O_{4} \to B a S O_{4} ↓ + 2 K C l$

$10$ mL of $1 M$ $B a C l_{2}$ and $5$ mL of $0.5 M$ $K_{2} S O_{4}$ are mixed together.
Here, $K_{2} S O_{4}$ is the limiting reagent.
The amount of $B a S O_{4}$ precipitated will be $\frac{5}{1000} \times 0.5 = 0.0025$ mol.

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Similar questions

10 ml of 1M Bacl2 solution and 5ml 0.5 M K2SO4 are mixed together to precipitate out BaS04 .The amount of BaSO4 precipited will be

a) 0.005.mol b) 0.00025mol

c)0.025mol. c)0.0025 mol

10

mL of

1 M

B a C l_{2}

solution and

5

mL of

0.5 M

K_{2} S O_{4}

are mixed together to precipitate out

B a S O_{4}

. The amount of

B a S O_{4}

precipated will be :

20

ml of

1 M B a C l_{2}

solution and

25

ml of

0.1

N a_{2} S O_{4}

are mixed to form

B a S O_{4}

. The amount of

B a S O_{4}

precipitated are:

Q. Calculate the amount of precipitate of

B a S O_{4}

formed when 30 mL of 0.1 M

B a C l_{2}

is mixed with 40 mL of 0.2 M

A l_{2} (S O_{4})_{3}

.
(Molar mass of barium = 137 g/mol, Molar mass of Al = 27 g/mol).

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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10 mL of 1 M BaCl2 solution and 5 mL of 0.5 M K2SO4 are mixed together to percipitate out BaSO4. The amount of BaSO4 precipated will be :

The correct option is C 0.0025 molThe reaction is as follows:BaCl2+K2SO4→BaSO4↓+2KCl10 mL of 1 M BaCl2 and 5 mL of 0.5 M K2SO4 are mixed together.Here, K2SO4 is the limiting reagent.The amount of BaSO4 precipitated will be 51000×0.5=0.0025 mol.

$10$ mL of $1 M B a C l_{2}$ solution and $5$ mL of $0.5 M K_{2} S O_{4}$ are mixed together to percipitate out $B a S O_{4}$ . The amount of $B a S O_{4}$ precipated will be :