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Mole Concept

The mass of B...

Question

The mass of $B a C O_{3}$ produced when excess $C O_{2}$ is bubbled through a solution of 0.205 mol $B a (O H)_{2}$ is
molar mass of Ba=137 $g m o l^{- 1}$

81 g

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40.40 g

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20.25 g

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162 g

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Solution

The correct option is B 40.40 g
$B a (O H)_{2} + C O_{2} \to B a C O_{3} + H_{2} O$
Atomic weight of $B a C O_{3} = 137 + 12 + 16 \times 3 = 197 g m o l^{- 1}$
Number of mole $= \frac{weight of substance}{molar weight}$
1 mole of $B a (O H)_{2}$ gives 1 mole of $B a C O_{3}$
$∴$ 205 mole of $B a (O H)_{2}$ will give 205 mole of $B a C O_{3}$
$∴$ weight of 0.205 mole of $B a C O_{3}$
$= number of mole \times molar mass of barium carbonate = 0.205 \times 197 = 40.38 g$

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Similar questions

Q. The mass of

B a C O_{3}

produced when excess of

C O_{2}

is bubbled through a solution of

0.205 mole B a (O H)_{2}

solution, is:
(Take molar mass of

B a = 137 g/mol

)

B a (O H)_{2} + C O_{2} \to B a C O_{3} + H_{2} O

Q. The mass of

B a C O_{3}

produced when excess of

C O_{2}

is bubbled through a solution of

0.205 mole B a (O H)_{2}

solution, is:
(Take molar mass of

B a = 137 g/mol

)

B a (O H)_{2} + C O_{2} \to B a C O_{3} + H_{2} O

Q. The mass of

B a C O_{3}

produced when excess

C O_{2}

is bubbled through a solution of 0.205 mol

B a (O H)_{2}

is
molar mass of Ba=137

g m o l^{- 1}

Q. The mass of

C l_{2}

produced in grams when 1A current is passed through NaCl solution for 30 minute is:

20 g

of a sample of

B a (O H)_{2}

is dissolved in

10 m L

0.5 N H C l

solution. The excess of

H C l

was titrated with

0.2 N N a O H .

The volume of

N a O H

required was

10 m L

. Calculate the percentage of

B a (O H)_{2}

in the sample.
(Given molar mass of

B a

137 g m o l^{- 1}

)

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Mole Concept

Standard XII Chemistry

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The mass of BaCO3 produced when excess CO2 is bubbled through a solution of 0.205 mol Ba(OH)2 is molar mass of Ba=137 gmol−1

The mass of $B a C O_{3}$ produced when excess $C O_{2}$ is bubbled through a solution of 0.205 mol $B a (O H)_{2}$ is
molar mass of Ba=137 $g m o l^{- 1}$