Question

A sample of CaCO3(s) is introduced into a sealed container of volume 0.638 L and heated to 1000 K until equilibrium is reached. The Kp for the reaction CaCO3(s)⇌CaO(s)+CO2(g) is 3.8×10−2 at this temperature. Calculate the mass of CaO(s) that is present at equilibrium.

Answer 1

Answer:

the mass of CaO present at equilibrium is, 0.01652g

Explanation:

$K_p = [CO_2]$ = 3.8×10⁻²

Now we have to calculate the moles of CO₂

Using ideal gas equation,

PV =nRT

P = pressure of gas = 3.8×10⁻²

T = temperature of gas = 1000 K

V = volume of gas = 0.638 L

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mole.k

$\frac{3.8 * 10^2 * 0.638}{0.0821 * 1000} \\= 2.95 * 10^-^4$

Now we have to calculate the mass of CaO

mass = 2.95 * 10 ⁻⁴ × 56

= 0.01652g

Therefore,

the mass of CaO present at equilibrium is, 0.01652g