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Hardness of Water and Treatment

Temporary har...

Question

Temporary hardness is due to bicarbonates of $M g^{2 +}$ and $C a^{2 +}$ . It is removed by addition of CaO as follows: $C a (H C O_{3})_{2} + C a O \to 2 C a C O_{3} + H_{2} O$ Mass of CaO required to precipitate $2 g C a C O_{3}$

A

2 g

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B

0.56 g

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C

0.28 g

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D

1.12 g

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Solution

The correct option is B 0.56 g
mole of $C a C O_{3} = \frac{2}{100} = 0.02$

From stoichiometry,
1 mole of $C a O$ is required to precipitate 2 moles of $C a C O_{3}$ and hence moles of $C a O$ required to precipitate 0.02 mole of $C a C O_{3}$
= $\frac{1}{2} \times 0.02 = 0.01$
so mass of CaO = $0.01 \times 56 = 0.56 g$

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Similar questions

Q. Temporary hardness is due to bicarbonates of

M g^{2 +}

C a^{2 +}

. It is removed by addition of CaO as follows:

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

Mass of CaO required to precipitate

2 g C a C O_{3}

Q. Temporary hardness is due to

H C O_{3}^{-}

M g^{2 +}

C a^{2 +}

. It is removed by addition of

C a O

C a (H C O_{3})_{2} + C a O ⟶ 2 C a C O_{3} + H_{2} O

C a O

required to precipitate

2

C a C O_{3}

Q. Temporary hardness is due to

H C O_{3}^{-}

M g^{2 +}

C a^{2 +}

. It is removed by addition of

C a O

.

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

C a O

(in g) required to produce

2

C a C O_{3}

precipitate is:

Following reaction is used to remove temporary hardness which is due to dissolved bicarbonates of

C a^{2 +}

M g^{2 +}

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

Actually lime (

C a O

) is added which in

H_{2} O

C a (O H)_{2}

and precipitates

C a C O_{3}

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

10

L of hard water requires

0.56

C a O

, temporary hardness of

C a C O_{3}

x \times 10^{2}

x

10

L of hard water with temporary hardness of

[C a (H C O_{3})_{2}]

0.56

g of lime. The reaction follows as:

C a (H C O_{3})_{2} + C a O ⟶ 2 C a C O_{3} + H_{2} O

.
Temporary hardness in terms of ppm of

C a C O_{3}

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