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Gram Equivalent weight of Phosphoric acid in the reaction, $2 H_{3} {PO}_{4} + Ca {(OH)}_{2} \to Ca {(H_{2} {PO}_{4})}_{2} + 2 H_{2} O$ is

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$\frac{98}{3} g$

$49 g$

$98 g$

$\frac{98}{4} g$

answer is C.

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Detailed Solution

$2 H_{3} {PO}_{4} + Ca {(OH)}_{2} \to Ca {(H_{2} {PO}_{4})}_{2} + 2 H_{2} O; H_{3} {PO}_{4} \to H_{2} {PO}_{4}^{-} Only one hydrogen atom is replaced . GEW = \frac{GMW}{1}$

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