Question

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?

Answer 1

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,
= $\frac{weight of CaCO3 (g)}{gram molecular weight}$
= $\frac{45}{100}$ = 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 = $\frac{nRT}{P}$.
Given that, P = 645 torr = 0.8487 atm (Since, 1 atm = 760 torr)
Therefore, V = $\frac{0.45 X 0.08206 X 800}{0.8487}$

= 34.8 l