Respuesta :
Explanation:
First, we will calculate the feed rate of alum as follows.
[tex]\frac{\text{60 mg alum}}{\text{1 L water}} \times \frac{\text{1000 L water}}{1 m^{3}} \times \frac{3800 m^{3}}{day} \times \frac{\text{1 g alum}}{\text{1000 mg alum}}[/tex]
= 228000 g/day
Converting this amount into g/min as follows.
[tex]\frac{228000 g}{1 day} \times \frac{1 day}{1440 min}[/tex]
= 158 g/min
Now, the chemical equation will be as follows.
[tex]Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O[/tex]
[tex]\frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{3 mmol SO^{2-}_{4}}}{\text{1 mmol alum}}[/tex]
= 0.151 mmol [tex]mmol SO^{2-}_{4}/L[/tex]
[tex]\frac{0.151 mmol SO^{2-}_{4}}{L} \times \frac{\text{2 meq SO^{2-}_{4}}}{\text{1 mmol SO^{2-}_{4}}} \times \frac{\text{1 meq Alk}}{\text{1 meq SO^{2-}_{4}}} \times \frac{\text{50 mg CaCO_{3}}}{\text{1 meq Alk}}[/tex]
= 15.15 mg [tex]CaCO_{3}[/tex]/L
For precipitate:
[tex]Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O[/tex]
[tex]\frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{2 mmol Al(OH)_{3}}}{\text{1 mmol alum}} \times \frac{\text{78 mg Al(OH)_{3}}}{\text{1 mmol Al(OH)_{3}}}[/tex]
= 7.88 [tex]Al(OH)_{3}/L[/tex]
[tex]\frac{7.88 mg Al(OH)_{3}}{1 L} \times \frac{3800 m^{3}}{1 day} \times \frac{1000 L}{1 m^{3}} \times \frac{1 kg}{10^{6} mg}[/tex]
= 29.9 [tex]Al(OH)_{3}/day[/tex]
