Question

250.1 g pure calcium carbonate are being heated and converted to solid calcium oxide and carbon dioxide. howmany liters of carbon dioxide at 88 C and 104.1 Kpa is released

Answer 1

CaCO3 (s) ---> CaO(s) + CO2(g)   is balanced equation

CaCO3 moles = mass of CaCO3 / molar mass of CaCO3 = ( 250.1 /100.07) = 2.501

CO2 moles produced = 2.501   ( since 1 CaCO3 gives 1 CO2)

T = 88C = 88+273 = 361 K ,

P = 104.1KPa = 104.1 x 0.009869   = 1.0274 atm ( since 1 KPa = 0.009869 atm)

we use PV = nRT equation to find Volume of gas CO2 ,

R is gas constant = 0.08206 liter atm/molK

now ( 1.0274 x V) = 2.501 x 0.08206 x 361

V = 72.1 Liters