Question

calculate the quantity of slaked lime required to soften 60000 litres of well water containing 16.2 gram of calcium bicarbonate per 100 litres​

Answer 1

Answer : The mass of slaked lime required is, 4440 grams.

Explanation :

First we have to calculate the mass of calcium bicarbonate.

As, 100 L of water contains 16.2 grams of calcium bicarbonate

So, 60,000 L of water contains [text]\frac{16.2g}{100L}\times 60,000L=9720g[/tex] of calcium bicarbonate

Now we have to calculate the Mass of slaked lime.

The balanced chemical reaction will be,

$Ca(HCO_3)_2+Ca(OH)_2\rightarrow 2CaCO_3+2H_2O$

From the balanced chemical reaction we conclude that,

1 mole of calcium bicarbonate react with 1 mole of slaked lime.

That means, $\frac{m_{Ca(HCO_3)_2}}{M_{Ca(HCO_3)_2}}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}$

where,

$m_{Ca(HCO_3)_2}$ = mass of calcium bicarbonate = 9720 g

$m_{Ca(OH)_2}$ = mass of slaked lime = ?

$M_{Ca(HCO_3)_2}$ = molar mass of calcium bicarbonate = 162 g/mole

$M_{Ca(OH)_2}$ = molar mass of slaked lime = 74 g/mole

Now put all the given values in this expression, we get the mass of slaked lime.

$\frac{9720g}{162g/mole}=\frac{m_{Ca(OH)_2}}{74g/mole}$

$m_{Ca(OH)_2}=4440g$

Therefore, the mass of slaked lime required is, 4440 grams.