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Ideal Gas Equation

When the calc...

Question

When the calcium carbonate is reacted with an excess amount of hydrochloric acid, how much $C a C O_{3}$ is required to produce $11.2$ liters of $C O_{2}$ at STP ?
(Given molar masses $:$ $C a C O_{3} = 100$ g/mol, $H C l = 36.5$ g/mol, $C O_{2} = 44.0$ g/mol)

25.0

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44.0

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50.0

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100.0

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none of the above

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Solution

The correct option is C $50.0$ g
$C a C O_{3} \to C a O + C O_{2}$
1 mole $C a C O_{3} =$ 1 mole $C O_{2}$ .
0.5 mole $C a C O_{3} =$ 0.5 mole $C O_{2}$ .
At STP, 1 mole of $C O_{2} =$ 22.4 L.
At STP, 0.5 mole of $C O_{2} = 0.5 \times 22.4 = 11.2$ L.
11.2 liters of $C O_{2}$ at STP $=$ 0.5 moles $C a C O_{3}$
1 mole $C a C O_{3} =$ 100. g.
0.51 mole $C a C O_{3} =$ 50.0 g.
Hence, 50.0 g $C a C O_{3}$ is required to produce 11.2 liters of $C O_{2}$ at STP.

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C a C O_{3} = 100

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C a C O_{3} \to C a O + C O_{2}

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