MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Percentage Composition

Temporary har...

Question

Temporary hardness is due to bicarbonates of $M g^{2 +} a n d C a^{2 +}$ , 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 $C a O$ required to precipitate 2 g $C a C O_{3}$ is:

A

2 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

0.56 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

0.28 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

1.12 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $0.56 g$
$C a (H C O_{3})_{2} + C a O \to 2 C a C O_{3} + H_{2} O$
Molecular mass of $C a O = 40 + 16 = 56 g$
Molecular mass of $C a O_{3} = 40 + 12 (16 \times 3) = 100 g$
Accoridng to the equation, $2$ moles of $C a C O_{3}$ i.e $200 g$ of $C a C O_{3}$ are formed from $56 g C a O$
$2 g$ of $C a C O_{3}$ will be formed from $= \frac{2 \times 56}{200} = 0.56 g r a m s$

Suggest Corrections

thumbs-up

0

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}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Percentage Composition and Molecular Formula

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Percentage Composition

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon