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Chapter 18: Problem 41

A first-stage recovery of magnesium from seawater is precipitation of \(\mathrm{Mg}(\mathrm{OH})_{2}\) with \(\mathrm{CaO};\) $$ \mathrm{Mg}^{2+}(a q)+\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{Ca}^{2+}(a q) $$ What mass of \(\mathrm{CaO}\), in grams, is needed to precipitate $1000 \mathrm{~kg}\( of \)\mathrm{Mg}(\mathrm{OH})_{2} ?$

Short Answer

Expert verified
Approximately 962.4 kg of CaO is needed to precipitate 1000 kg of Mg(OH)₂.

Step by step solution

01

Determine the stoichiometric ratio between Mg(OH)₂ and CaO

In the balanced chemical equation: Mg²⁺(aq) + CaO(s) + H₂O(l) -> Mg(OH)₂(s) + Ca²⁺(aq) It can be observed that for each mole of Mg(OH)₂ produced, one mole of CaO is needed. The stoichiometric ratio between Mg(OH)₂ and CaO is 1:1.
02

Calculate the moles of Mg(OH)₂ to be precipitated

Given that we want to precipitate 1000 kg of Mg(OH)₂, we should first convert this mass to moles. The molar mass of Mg(OH)₂ is: Molar mass of Mg(OH)₂ = 24.31 (Mg) + (2 * 16.00 + 2 * 1.01)= 58.32 g/mol Now, we can convert the mass of Mg(OH)₂ to moles: Moles of Mg(OH)₂ = (1000 kg * 1000 g/kg) / 58.32 g/mol ≈ 17160 moles
03

Calculate the moles of CaO required

As determined in Step 1, the stoichiometric ratio between Mg(OH)₂ and CaO is 1:1. Therefore, the moles of CaO required is the same as the moles of Mg(OH)₂: Moles of CaO = 17160 moles
04

Calculate the mass of CaO required

Now, we can convert the moles of CaO to mass using its molar mass. The molar mass of CaO is: Molar mass of CaO = 40.08 (Ca) + 16.00 (O) = 56.08 g/mol We can now calculate the mass of CaO required: Mass of CaO = moles of CaO * molar mass of CaO Mass of CaO = 17160 moles * 56.08 g/mol ≈ 962432 g So, approximately 962432 g (or 962.4 kg) of CaO is needed to precipitate 1000 kg of Mg(OH)₂.

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