A small amount of $C a C O_{3}$ completely neutralizes 525 ml of 0.1 N HCI and no acid is left at the end. After converting all calcium chloride to $C a S O_{4},$ how much plaster of Paris can be obtained?

1.916 g

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5.827 g

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3.57 g

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3.81 g

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Solution

The correct option is C 3.81 g
We have reaction
$C a C O_{3} + 2 H C l \to C a C l_{2} + C O_{2} + H_{2} O$
1 2
26.25 52.5 (0.1 x 525)

Also, $C a C l_{2} + H_{2} S O_{4} \to C a S O_{4} + 2 H C l$
millimole 26.25

$C a S O_{4} + \frac{1}{2} H_{2} O \to C a S O_{4} . \frac{1}{2} H_{2} O \Leftarrow$ plaster of paris
millimole 26.25

M. wt. of plaster of paris $(C a . S O_{4} . \frac{1}{2} H_{2} O) = 145$ gm/mol

Moles of plaster of paris $= 26.25 \times 10^{- 3} = 3.806 g m = 3.81 g m$

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Similar questions

Plaster of Paris is prepared by heating one of the following to a temperature of $100^{o} C$ . This is :

(a) $C a S O_{3} 2 H_{2} O$

(b) $C a C l_{2} 2 H_{2} O$

(d) $C a S O_{4} .2 H_{2} O$

C a C O_{3}

525

\frac{N}{10} H C l

C a S O_{4}

C a C O_{3}

525 m L

0.1 N H C l

C a S O_{4}

C a S O_{4}

H_{2} O

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