What volume of co2 gas at 645 torr and 800. k could be produced by the decomposition of 45.0 g of caco3? caco3(s) → cao(s) + co2(g) (r = 0.08206 l • atm/k • mol?
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]
