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Standard XII

Chemistry

Stoichiometric Calculations

What is the w...

Question

What is the weight of $C a (O H)_{2}$ required for $10$ litres of water to remove temporary hardness of $100$ ppm due to $C a (H C O_{3})_{2}$ ?

1.62

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0.74

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7.4

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None of the above

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Solution

The correct option is A $0.74$ g
$10^{6}$ g water contain $100$ g $C a C O_{3}$ .
$10$ L or $10^{4}$ g water contain $\frac{100}{10^{6}} \times 10^{4} = 1$ g of $C a C O_{3}$
$100$ g $C a C O_{3} = 162$ g of $C a (H C O_{3})_{2}$
or $1$ g $C a C O_{3} = 1.62$ g of $C a (H C O_{3})_{2}$
or $1$ g $C a C O_{3}$ $= 0.01$ mol
$C a (H C O_{3})_{2} + C a (O H)_{2} \to 2 C a C O_{3} ↓ + 2 H_{2} O$
Moles of $C a (O H)_{2}$ required $= 0.01$
Weight of $C a (O H)_{2}$ required $= 0.74$ g

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Similar questions

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

C a^{2 +}

and

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

forms

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

g of

C a O

, temporary hardness of

C a C O_{3}

x \times 10^{2}

ppm. Find

x

Q. Hardness of water is measured in terms of ppm (parts per million) of

C a C O_{3}

. It is the amount (in g) of

C a C O_{3}

present in

10^{6} g H_{2} O

. In a sample of water,

10

L required

0.56

g of

C a O

to the remove temporary hardness of

H C O_{3}^{-}

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

Temporary hardness is :

Q. What is the amount of lime

C a {(O H)}_{2}

required to remove the hardness in

60

litres of pond water containing

1.62

mg of calcium bicarbonate per

100

mL of water?

Q. What is the weight of

C a O

required to remove hardness of

10^{6} L

of water containing

1.62 g o f C a (H C O_{3})_{2}

1 L

?
The reaction is ,

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

Q. Calculate the weight of

C a O

required to remove the hardness of 1000000 litre of water containing 1.62 g of

C a C O_{3}

per litre.

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What is the weight of Ca(OH)2 required for 10 litres of water to remove temporary hardness of 100 ppm due to Ca(HCO3)2?

The correct option is A 0.74 g106 g water contain 100 g CaCO3.10 L or 104 g water contain100106×104=1 g of CaCO3100 g CaCO3=162 g of Ca(HCO3)2or 1 g CaCO3=1.62 g of Ca(HCO3)2or 1 g CaCO3 =0.01 molCa(HCO3)2+Ca(OH)2→2CaCO3↓+2H2OMoles of Ca(OH)2 required=0.01Weight of Ca(OH)2 required=0.74 g

What is the weight of $C a (O H)_{2}$ required for $10$ litres of water to remove temporary hardness of $100$ ppm due to $C a (H C O_{3})_{2}$ ?