What mass of caso3 must have been present initially to produce 14.5 l of so2 gas at a temperature of 12.5°c and a pressure of 1.10 atm?

When the reaction equation is:

CaSO3(s) → CaO(s) + SO2(g)

we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.

to get the moles of SO2 we are going to use the ideal gas equation:

PV = nRT

when P is the pressure =  1.1 atm

and V is the volume = 14.5 L 

n is the moles' number (which we need to calculate)

R ideal gas constant = 0.0821

and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K

so, by substitution:

1.1 * 14.5 L = n * 0.0821 * 285.5

∴ n = 1.1 * 14.5 / (0.0821*285.5)

       = 0.68 moles SO2

∴ moles CaSO3 = 0.68 moles

so we can easily get the mass of CaSO3:

when mass = moles * molar mass

and we know that the molar mass of CaSO3= 40 + 32 + 16 * 3 = 120 g/mol

∴ mass = 0.68 moles* 120 g/mol = 81.6 g