What is the weight of ca(oh)2 required for 10 litre of water remove temporary hardness of 100 ppm due to ca(H03)2

Question

The answer 0.74

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Answer 1

1 mol Ca(OH)2 removes 1 mol Ca(HCO3)2; therefore, to remove 100 ppm harness (as CaCO3) will require (100 x 74/100) = 74 ppm Ca(OH)2.
(Note: To convert harness in ppm CaCO3 to Ca(OH)2, use the coefficients in the balanced equation. That is a 1:1 mole ratio in which 1 mol Ca(OH)2 = 1 mol CaCO3; therefore, 100 ppm CaCO3 x (molar mass Ca(OH)2/molar mass CaCO3) = 100ppm x 74/100 = 74 mg Ca(OH)2.
74 ppm = 74 mg/L = 740 mg/10L = 0.74 g/10L.

Answer 2

10
6
g water contain 100 g CaCO
3
​
.
10 L or 10
4
g water contain
10
6

100
​
×10
4
=1 g of CaCO
3
​

100 g CaCO
3
​
=162 g of Ca(HCO
3
​
)
2
​

or 1 g CaCO
3
​
=1.62 g of Ca(HCO
3
​
)
2
​

or 1 g CaCO
3
​
=0.01 mol
Ca(HCO
3
​
)
2
​
+Ca(OH)
2
​
→2CaCO
3
​
↓+2H
2
​
O
Moles of Ca(OH)
2
​
required=0.01
Weight of Ca(OH)
2
​
required=0.74 g