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The quantity of alum required to treat 18 million liters of water per day is 252 kg/day. Calculate the amount of CO2 released (in kg/day) given the reaction:

\[ \text{Al}_2(\text{SO}_4)_3 \cdot 18\text{H}_2\text{O} + 3\text{Ca}(\text{HCO}_3)_2 \rightarrow 2\text{Al(OH)}_3 + 3\text{CaSO}_4 + 18\text{H}_2\text{O} + 6\text{CO}_2 \]

A. 50.4
B. 10.1
C. 100.8
D. 20.2
E. None of these
F. Not enough information

Answer :

Without additional information such as the molar masses of the compounds, it is impossible to accurately determine the amount of [tex]CO2[/tex] released. The correct answer is option e.

The question deals with a chemical reaction involving Alum and Calcium bicarbonate. The quantity of [tex]CO2[/tex] released depends on the reaction stoichiometry, not the amount of alum used directly.

The subject of the question deals with a chemical reaction involving Alum [tex](Al2(SO4)3.18H2O)[/tex] and Calcium bicarbonate [tex](3Ca(HCO3)2)[/tex].

From the reaction, we can see that 6 moles of carbon dioxide ([tex]CO2[/tex]) are produced. The quantity of alum used does not directly determine the quantity of[tex]CO2[/tex] released, as it is the reaction stoichiometry that tells us that 1 mole of alum reacts to produce 6 moles of [tex]CO2[/tex].

However, without knowing the molar masses of the compounds and being given the quantity of alum in moles, we cannot calculate the exact amount of [tex]CO2[/tex] released. Therefore, the correct answer would be option e 'Not enough information'.

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Complete Question

The quantity of Alum required to treat 18 million liters of water per day is 252 kg/d. The amount of CO2 released (kg/day) is Al2(SO4)3.18H,0 + 3Ca(HCO3)2 2Al(OH), + 3CaSO4 + 18H,0 + 6CO2 50.4

O a. 10.1

O b. 100.8

O c. 20.2

O d. None of these

O e. Not enough information

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