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Standard XII

Chemistry

Carbon Dioxide

A 10.0 g sa...

Question

A $10.0$ g sample of a mixture of $C a C l_{2}$ and $N a C l$ is treated to precipitate all the calcium as calcium carbonate. This $C a C O_{3}$ is heated to convert all the $C a$ to $C a O$ and the final mass of $C a O$ is $1.62$ g. If $X$ percentage by mass of $C a C l_{2}$ is in the original mixture, then value of $10 X$ is :

321

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200

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380

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None of these

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Solution

The correct option is A $321$
$C a C l_{2} 1 m o l \to C a C O_{3} 1 m o l \to a O 1 m o l$
Let the $10$ g mixture contain $x$ g of $C a C l_{2}$ .
Moles of $C a C l_{2} = \frac{x}{111}$ , while Moles of

$C a O = \frac{1.62}{56}$ .

$1$ mole of $C a C l_{2}$ produces $1$ mole of $C a O$ .
$∴ \frac{x}{111} = \frac{1.62}{56}$
$∴ x = 3.21$ g.

$10 x = 32.1 \times 10$ = 321

$%$ of $C a C l_{2} = \frac{3.21}{10} \times 100 = 32.1 %$

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A 10.0 g sample of a mixture of CaCl2 and NaCl is treated to precipitate all the calcium as calcium carbonate. This CaCO3 is heated to convert all the Ca to CaO and the final mass of CaO is 1.62 g. If X percentage by mass of CaCl2 is in the original mixture, then value of 10X is :

The correct option is A 321CaCl21mol→CaCO31mol→aO1molLet the 10 g mixture contain x g of CaCl2.Moles of CaCl2=x111, while Moles of CaO=1.6256.1 mole of CaCl2 produces 1 mole of CaO.∴x111=1.6256∴x=3.21 g.10x=32.1×10 = 321% of CaCl2=3.2110×100=32.1%