Hydrated calcium sulphate when heated at 120- $130^{0} C$ loses ______.

1 / 2 H_{2} O

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H_{2} O

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3 / 2 H_{2} O

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H_{2} O

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Solution

The correct option is A $3 / 2 H_{2} O$
Hydrated calcium sulphate (gypsum) loses three-fourth of its water of crystallisation and forms the hemihydrate known as plaster of paris.
$C a S O_{4} \cdot 2 H_{2} O - ---- \to 120^{0} C C a S O_{4} \cdot \frac{1}{2} H_{2} O + 1 \frac{1}{2} H_{2} O M o n o c l i n i c P l a s t e r g y p s u m o f p a r i s$

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Similar questions

Complete the following reaction as balanced chemical equations:1.Na+H2O=? 2.K+H2O=? 3.Ca+H2o?

C u + H_{2} S O_{4} \to C u S O_{4} + {S O}_{2} ↑ + H_{2} O

F e + H_{2} O \to {F e}_{3} O_{4} + H_{2} ↑

{C H}_{4} + O_{2} \to C O_{2} + H_{2} O

C u + H_{2} {S O}_{4} - \to C u S O_{4} + {S O}_{2} ↑ + H_{2} O

F e + + H_{2} O - \to {F e}_{3} O_{4} + H_{2} ↑

{C H}_{4} + O_{2} \to C O_{2} + H_{2} O

H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H = - 68.09 k c a l

K + H_{2} O + w a t e r \to K O H (a q .) + \frac{1}{2} H_{2}; Δ H = - 48.0 k c a l

K O H + w a t e r \to K O H (a q); Δ H = - 14.0 k c a l

K O H

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