Explanation:
The given reaction is,
[tex]CaCO_{3}->CaO+CO_{2}[/tex]
let [tex]n[/tex] be the number of moles of [tex]CaCO_{3}[/tex]
For one mole of [tex]CaCO_{3}[/tex],one mole of [tex]CO_{2}[/tex] is formed.
[tex]\text{number of moles}=\frac{\text{given weight}}{\text{molar weight}}[/tex]
[tex]\text{molar weight}[/tex] of [tex]CaCO_{3}[/tex]=100g
Given that [tex]given weight=14g[/tex]
[tex]n=\frac{14}{100}[/tex]
So,number of moles of [tex]CO_{2}=n=0.14[/tex]
using [tex]PV=nRT[/tex] with [tex]P=\text{1 atm,}T=\text{1000K,}R={\text{0.0821 L atm }mol^{-1}K^{-1}[/tex]
[tex]v=\frac{nRT}{P}=0.14\times 0.0821\times 1000\times \frac{1}{1} =11.4L[/tex]
