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Seawater cont...

Question

Seawater contains 1272 g of $M g^{2 +}$ per metric ton (1 megagram). The amount of slaked lime (in Kg) must be added to 1.0 metric ton of seawater to precipitate all of the $M g^{2 +}$ ion is _____.

[Atomic weight of Mg =24 and Ca =40]. Write the nearest integer.

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Solution

The reaction is: $M g^{2 +} + C a (O H)_{2} \to M g (O H)_{2} + C a^{2 +}$

24 g 74 g

1272 g ?

Since 24g of $M g^{2 +}$ is precipitated by 74g of $C a (O H)_{2}$ .

Therefore, 1272g of $M g^{2 +}$ is precipitated by $\frac{74 \times 1272}{24} g$

$= 3922 g = 3.922 k g \approx 4 k g$ .

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