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A sample of calcium carbonate is 80% pure. 25 g of this sample is treated with excess of HCl. How much volume of ${CO}_{2}$ will be obtained at N.T.P.?
${CaCO}_{3} + 2 HCl \to {CaCl}_{2} + H_{2} O + {CO}_{2}$

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5.6 L

11.2 L

4.48 L

2.24 L

answer is C.

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Detailed Solution

$C a C O_{3} + 2 H C l \to C a C l_{2} + H_{2} O + C O_{2} ↑$ $(\begin{array}{l} 25 g = 100 % \\ ? - - - 80 % \end{array})$

$100 gr {CaCO}_{3} ---------- 22.4 L {CO}_{2} ↑$ $\frac{80 \times 25}{100} = 20 g r$

$20gr {CaCO}_{3} --------------- ?$

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