What will be the volume of CO 2 at NTP obtained on heating 10 grams of 90 - pure limestoneA- 22-4 litreB- 2-016 litreC- 2-24 litreD- 20-16 litre

The correct option is B 2.016 litreCaCO_3 → CaO + CO_2 10 gm 90% pure 9gm = 9/100 mole CaCO_3 ≡ CO_2 = 0.09 mole At NTP Vol.CO_2=0.09 × 22.4 = 2.016 L ...

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Standard XII

Chemistry

Stoichiometric Calculations

What will be ...

Question

What will be the volume of $C O_{2}$ at NTP obtained on heating 10 grams of ( $90 %$ pure) limestone

22.4 litre

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2.016 litre

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2.24 litre

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20.16 litre

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Solution

The correct option is B 2.016 litre
$C a C O_{3} \to C a O + C O_{2}$
10 gm
$90 %$ pure 9gm $= \frac{9}{100} m o l e$
$C a C O_{3} \equiv C O_{2} = 0.09 m o l e$
At NTP Vol. $C O_{2} = 0.09 \times 22.4 = 2.016 L$

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Q. What will be the volume of

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Calcium carbonate decomposes on heating according to the equation:

$C a C O_{3}$ (s) $⟶$ CaO(s)+ $C O_{2}$

At STP, the volume of $C O_{2}$ obtained by thermal decomposition of 50g of $C a C O_{3}$ will be:

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What will be the volume of CO2 at NTP obtained on heating 10 grams of (90% pure) limestone

The correct option is B 2.016 litreCaCO3→CaO+CO2 10 gm 90% pure 9gm =9100 mole CaCO3≡CO2=0.09 mole At NTP Vol.CO2=0.09×22.4=2.016 L

What will be the volume of $C O_{2}$ at NTP obtained on heating 10 grams of ( $90 %$ pure) limestone

The correct option is B 2.016 litre
$C a C O_{3} \to C a O + C O_{2}$
10 gm
$90 %$ pure 9gm $= \frac{9}{100} m o l e$
$C a C O_{3} \equiv C O_{2} = 0.09 m o l e$
At NTP Vol. $C O_{2} = 0.09 \times 22.4 = 2.016 L$