47.43 kg of Zeolite A is needed to soften a week's supply of the water.
<h3>What is Zeolites ?</h3>
Zeolites are 3-dimensional crystalline solids which are formed either synthesized or can be naturally. Zeolites belongs to class microporous solids.
Given
Hard water supply per day = 25,000L
Ca⁺² concentration = 0.0045 mol/L
Total mole of Ca⁺² = 25,000 × 0.0045
= 112.5 mol
Mass of Ca⁺² in hard water = Mole × Molar mass
= 112.5 × 40.08 g
= 4509 g
Concentration of Mg⁺² = 0.00092 mol/L
Total mole of Mg⁺² = 0.00092 × 25000
= 23
Mass of Mg⁺² in Hard water = Mole × Molar mass
= 23 g × 18.04
= 414.92 g
Molar mass of Zeolite = 2190g/mol
No of mole of Ca⁺² removed by 1 mole of Zeolite A = 6
Mass of Ca⁺² removed by 2190 g of Zeolite A = 240.48g
Mass of Zeolite A required to remove 4509 g of Ca⁺² = 33.73kg
Assume efficiency of Zeolite is 85%
To remove Ca⁺² the mass of Zeolite A = 39.68kg
No of mole of Mg⁺² removed by 1 mole of Zeolite A = 6
Mass of Mg⁺² removed by 2190g of Zeolite A = 145.86g
To remove 414.92 g of Mg⁺² the mass of Zeolite = 6.59kg
Assume the 85% as the efficient of Zeolite A
Mass of Zeolite A required to remove Mg⁺² = 7.75kg
Total mass of Zeolite A = 39.68kg + 7.75kg
= 47.43kg
Thus from the above conclusion we can say that 47.43 kg of Zeolite A is needed to soften a week's supply of the water.
Learn more about the Zeolites here: brainly.com/question/14976531
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