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Stoichiometric Calculations

To get 5.6 ...

Question

To get $5.6$ L of $C O_{2}$ at STP, weight of $C a C O_{3}$ to be decomposed is:

100

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50

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25

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75

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Solution

The correct option is B $25$ g
The reaction is $C a C O_{3} \to C a O + C O_{2}$ .

One mole of calcium carbonate forms one mole of carbon dioxide.

At STP, 5.6 L of carbon dioxide corresponds to 0.25 moles.

They will be obtained from 0.25 moles of calcium carbonate.

Molar mass of $C a C O_{3} = 100 g$
0.25 moles of calcium carbonate (molecular weight 100 g/mole) corresponds to 25 g.

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