Find the equivalent weight of $H_{3} P O_{4}$ in the following reaction:
$C a (O H)_{2} + H_{3} P O_{4} \to C a H P O_{4} + 2 H_{2} O$

27

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60

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12

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49

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Solution

The correct option is A $49$
The balanced reaction is as follows:

$C a (O H)_{2} + H_{3} P O_{4} \to C a H P O_{4} + 2 H_{2} O$
Two hydrogen atoms of $H_{3} P O_{4}$ have been replaced.

$Equivalent weight = \frac{Molecular weight}{Replaced H} = \frac{98}{2} = 49$ g

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Similar questions

The equivalent weight of H3PO4 in following reaction

H3PO4 + Ca(OH)2 ---> CaHPO4 +2 H2O

{H_{3} P O}_{4}

{H_{3} P O}_{4} + C a {(O H)}_{2} ⟶ {C a H P O}_{4} + 2 H_{2} O

C a (O H)_{2} + H_{3} P O_{4} \to C a H P O_{4} + 2 H_{2} O

H_{3} P O_{4}

H_{3} P O_{4}

C a (O H)_{2} + H_{3} P O_{4} \to C a H P O_{4} + 2 H_{2} O

(C a = 40, P = 31, O = 16)

C a (O H)_{2} + H_{3} P O_{4} \to C a H P O_{4} + 2 H_{2} O

H_{3} P O_{4}

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