From the reaction: caco3(s) → cao(s) + co2(g) it can be seen that,
1 mol (i.e. 100 g) of CaCO3 gives 1 mol (i.e. 44 g) of CO2
Now, number of moles of CaCO3 present in reaction system,
= [tex] \frac{weight of CaCO3 (g)}{gram molecular weight} [/tex]
= [tex] \frac{45}{100} [/tex] = 0.45 mol
So, 0.45 mol of CaCO3 will give 0.45 mol of CO2.
From ideal gas equation, we know that PV = nRT
V = [tex] \frac{nRT}{P} [/tex].
Given that, P = 645 torr = 0.8487 atm (Since, 1 atm = 760 torr)
Therefore, V = [tex] \frac{0.45 X 0.08206 X 800}{0.8487} [/tex]
= 34.8 l
