Question

Calculate the amount of lime, $Ca(OH)_{2}$, required to remove hardness of 50,000 litres of well water which has been found to contain 1.62 g of calcium bicarbonate per 10 litre.

Answer 1

Given,  

10 L of well water contains 1.62 gram of Calcium Bicarbonate $Ca{ \left( { HCO }_{ 3 } \right) }_{ 2 }$

Therefore, totally 50,000 L of well water contains 8100 of $Ca{ CO }_{ 3 }$

We know that,

Molar Mass of Calcium Bicarbonate $Ca{ \left( { HCO }_{ 3 } \right) }_{ 2 } = 162 g$

Molar Mass of lime water $Ca{ \left( OH \right) }_{ 2 } = 74 g$

$Ca{ \left( { HCO }_{ 3 } \right) }_{ 2 }\quad +\quad Ca{ \left( OH \right) }_{ 2 }\quad \rightarrow \quad 2Ca{ CO }_{ 3 }\quad +\quad 2{ H }_{ 2 }O$

Then,

Number of moles of $Ca{ \left( { HCO }_{ 3 } \right) }_{ 2 }$ = Number of moles of $Ca{ \left( OH \right) }_{ 2 }$

$\frac { 8100 }{ 162 } \quad =\quad \frac { m }{ 74 }$

$m\quad =\quad \frac { 74\quad \times \quad 8100 }{ 162 }$

m = 3700 g

Therefore, the required mass of lime is 3700 g.

Answer 2

Given,  

10 L of well water contains 1.62 gram of Calcium Bicarbonate Ca{ \left( { HCO }_{ 3 } \right) }_{ 2 }Ca(HCO3​)2​

Therefore, totally 50,000 L of well water contains 8100 of Ca{ CO }_{ 3 }CaCO3​

We know that,

Molar Mass of Calcium Bicarbonate Ca{ \left( { HCO }_{ 3 } \right) }_{ 2 } = 162 gCa(HCO3​)2​=162g

Molar Mass of lime water Ca{ \left( OH \right) }_{ 2 } = 74 gCa(OH)2​=74g

Ca{ \left( { HCO }_{ 3 } \right) }_{ 2 }\quad +\quad Ca{ \left( OH \right) }_{ 2 }\quad \rightarrow \quad 2Ca{ CO }_{ 3 }\quad +\quad 2{ H }_{ 2 }OCa(HCO3​)2​+Ca(OH)2​→2CaCO3​+2H2​O

Then,

Number of moles of Ca{ \left( { HCO }_{ 3 } \right) }_{ 2 }Ca(HCO3​)2​ = Number of moles of Ca{ \left( OH \right) }_{ 2 }Ca(OH)2​

\frac { 8100 }{ 162 } \quad =\quad \frac { m }{ 74 }1628100​=74m​

m\quad =\quad \frac { 74\quad \times \quad 8100 }{ 162 }m=16274×8100​

m = 3700 g

Therefore, the required mass of lime is 3700 g.