Question

e) calculate the amount of lime (84% pure) and soda (92%) needed in kg for softening 3,00,000 litre of
water using 32.8 ppm of NaAlO, as a coagulant the impurities present in water are as follows: Ca**
= 240mg /1, Mg** = 96 mg/1, HCO = 732 mg/l, dissolve CO2 - 44 mg /1, NaCl 60 mg/l and Fe 0,
160 mg/l.​

Answer 1

Answer:

Explanation:

LIME = 74/100 (5+75+40+2*10+5) * (1million/106) * (100/90[purity of water])

= 119.2 kg

SODA= 106/100(30+40+5-75)1100/98

= 0 kg

Answer 2

Answer:

Permanent hardness $=\mathbf{7 5 p p m}$

Explanation:

Step 1: Temporary hardness is cause by the bicarbonates of calcium and magnesium where as permanent hardness is caused by the sulfate , nitrate and chloride of the calcium and magnesium.Temporary hardness can be removed by boiling and permanent hardness can be removed by softening techniques.

Lime required

$\text { Lime }=\frac{74}{100}(25.15+(2 \times 25)+25) \times \frac{20000}{10^5} \times \frac{100}{84} \text { Lime }=1.7619 \mathrm{~K} g$

Soda required

$\text { Soda }=\frac{106}{100}(25+25+25) \times \frac{20000}{10^{\circ}} \times \frac{100}{92} \text { Soda }=2.88 K g$

Step 2: Temporary hardness

$\text { T. } \mathrm{H}=\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2+\mathrm{Mg}\left(\mathrm{HCO}_3\right)_2=25+25$

Temporary hardness $=\mathbf{5 0 p p m}$

Permanent Hardness

$\begin{aligned}\text { P. } M & =\mathrm{CaSO}_4+\mathrm{CaCl}_2+\mathrm{MgSO}_4 \\& =25+25+25\end{aligned}$

Permanent hardness $=\mathbf{7 5 p p m}$.

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