Question

The volume of CO2 at NTP obtained by complete decomposition of 1gm of marble is :

Answer 1

Your answer goes like this....

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As we know that marble=Calcium carbonate (CaCo₃).

100 grams of calcium carbonate gives 24 dm₃ of Carbon dioxide at Normal temperature and pressure(NTP).

⇒1 gram of caco₃ gives 24÷100⇒0.24

   Hence,The volume of CO2 at NTP obtained by complete decomposition of 1gm of marble is "0.24"

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Thank you!

Answer 2

The volume of $CO_2$ at NTP obtained by complete decomposition of 1gm of marble is 0.24 L

Explanation:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

According to stoichiometry:

100 g of $CaCO_3$ produces = 1 mole of $CO_2$

1 g of $CaCO_3$ produces=$\frac{1}{100}\times 1=0.01$ moles of $CO_2$

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

$PV=nRT$

where,

P = pressure of the gas  = 1 atm (at NTP)

V = volume of the gas  = ?

T = temperature of the gas  = 293 K (at NTP)

n = number of moles of gas = 0.01

R = Gas constant= 0.0821 Latm/Kmol

$1atm\times V=0.01\times 0.0821Latm/Kmol\times 293K$

$V=0.24L$

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