A sample of calcium carbonate is $80$ % pure, $25$ g of this sample is treated with excess of $H C l$ . How much volume of $C O_{2}$ will be obtained at $1$ atm and $273$ K?

4.48

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3.44

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6.34

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5.87

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Solution

The correct option is A $4.48$ L
$C a C O_{3} + 2 H C l \to C a C l_{2} + C O_{2} + H_{2} O$

$100$ g of $C a C O_{3}$ gives $22.4$ L of $C O_{2}$ gas.

So, $25 \times 0.8 = 20$ g of $C a C O_{3}$ will give $4.48$ L $C O_{2}$ gas.

Hence, option A is correct.

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