Question

A sample of CaCO3(s) is introduced into a sealed container of volume 0.821 litre and heated to 1000 K until equilibrium is reached. The equilibrium constant for the reaction Caco,(s)= CaO(s)+CO, (g) is 4 x 10-2 atm at this temperature. Calculate the mass of Cao present at equilibrium. ​

Answer 1

Answer : The mass of CaO present at equilibrium is, 0.0224 grams.

Explanation :

The balanced chemical reaction is,

$CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$

The expression of equilibrium constant for the given reaction will be,

$K_p=(p_{CO_2})$

Now put the value of $K_p$, we get

$(p_{CO_2})=4\times 10^{-2}atm$

Now we have to calculate the moles of $CO_2$.

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of gas = $4\times 10^{-2}atm$

V = volume of gas = 0.821 L

T = temperature of gas = 1000 K

n = number of moles of gas = ?

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

Now put all the given values in the ideal gas equation, we get the number of moles of gas.

$(4\times 10^{-2}atm)\times (0.821L)=n\times (0.0821L.atm/mole.K)\times (1000K)$

$n=4\times 10^{-4}mole$

Now we have to calculate the moles of CaO.

From the balanced chemical reaction, we conclude that

The mole ratio of $CaO$ and $CO_2$ is, 1 : 1

So, the moles of CaO = moles of $CO_2$ = $4\times 10^{-4}mole$

Now we have to calculate the mass of CaO.

$\text{Mass of }CaO=\text{Moles of }CaO\times \text{Molar mass of }CaO$

$\text{Mass of }CaO=4\times 10^{-4}mole\times 56g/mole=0.0224g$

Therefore, the mass of CaO present at equilibrium is, 0.0224 grams.

Answer 2

Answer:

22.4 mg

here's the answer and hope this helps you.