A sample of hard water contains 96 ppm of $S O_{4}^{2 -}$ and 183 ppm of $H C O_{3}^{-},$ with $C a^{2 +}$ as only cation. The number of moles of CaO required to remove $H C O_{3}^{-} & S O_{4}^{2 -}$ from 1000 kg of this water is $x & y$ respectively then the value of $10 x + y$ is

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answer is 15.

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Detailed Solution

$n H C O_{3}^{-} = \frac{183}{61} = 3$

$n S O_{4}^{2 -} = \frac{96}{96} = 1$

$C a O + 2 H C O_{3}^{-} \to C a C O_{3} + H_{2} O + C O_{3}^{2 -}$

2 moles of $H C O_{3}^{-} \to 1$ mole of CaO

3 Moles $\to ?$

3/2 = 1.5

$C a O + S O_{4}^{2 -} + H_{2} O \to C a S O_{4} + 2 O H^{-}$

1 mole of Cao = 1 mole of $S O_{4}^{2 -}$

$x = 1.5, y = 010 x + y = 15 + 0 = 15$

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