Question

A sample of CaCO3(s) is introduced into a sealed container of volume 0.674 L and heated to 1000 K until equilibrium is reached. The Kp for the reaction CaCO3(s)⇌CaO(s)+CO2(g) is 3.9×10−2 at this temperature. Calculate the mass of CaO(s) that is present at equillibrium.

please show work

Answer 1

Firstly write done the all things mentioned in the question.

V= 0.674L

T =1000K

Kp = pressure CO2 = 3.9 x 10^^-2

R = gas constant = 0.08206

We know the Ideal gas law pV= nRT

hence n= pV/RT
moles CO2 = pV/RT = 3.9 x 10^-2 x 0.654 / 0.08206 x 1000=0.000311

The number of moles of CO2 = Number of moles of CaO ( stiochemetric coefficient from the given reaction. It is 1:1)

hence moles CaO =0.000311

we know that

number of moles = weight of sample / Equivalent weight

molar mass of CaO = 56.077g/mol

hence

mass of CaO= moles x Equvalent weight(molecular weight)

mass CaO = 0.000311 mol x 56.077 g/mol=0.0174 g

Answer: mass of CaO = 0.0174 g