Question

A 2.00(g) sample of a mixture of CaCO(s) and BaO(s) is placed in a 1.00 L container containing CO₂(g) at a pressure of 785 torr and a T = 525 C. The CO₂(g) reacts with the solids forming calcium carbonate and barium carbonate to completion. The pressure of the CO₂(g) after the reaction is complete is 964 Pa. Write the chemical reactions and calculate the mole fraction X of CaO and BaO. Please show all work.

Answer 1

Ans:

Mass of the mixture = 2 g

Let ‘x’ g be the mass of CaO in the mixture

Then mass of BaO in the mixture = (2-x)g

Molar mass of CaO = 56.1 g/mol

Molar mass of BaO = 153.3 g/mol

No. of moles of CaO taken = (x/56.1) mol

No. of moles of BaO taken = {(2-x)/153.3} mol

Volume of the container, V = 1 L = 10^-3 m³

Temperature, T = 525 ⁰C = 798 K

Pressure, P = 785 torr = 104658 Pa (1 torr = 133.322 Pa)

n = no. of moles of gas

R = gas constant = 8.314 m³⋅Pa⋅K⁻¹⋅mol⁻¹

According to Ideal gas equation:

PV = nRT & n = PV/RT

n = {(104658 Pa x 10^-3 m³)/( 8.314 m³⋅Pa⋅K⁻¹⋅mol⁻¹ x 798 K)} = 0.015775 mol

No. of moles of CO2 taken initially = 0.015775 mol

==> Pressure of CO2 after the reaction = 964 Pa

According to Ideal gas equation:

PV = nRT & n = PV/RT

n = {(964 Pa x 10^-3 m³)/( 8.314 m³⋅Pa⋅K⁻¹⋅mol⁻¹ x 798 K)} = 0.000145 mol

No of moles of CO2 reacted = (0.015775 mol - 0.000145 mol) = 0.01563 mol

Reaction of CaO & BaO with CO2:

CaO + CO2 --> CaCO3 (1 mol of CaO reacts with 1 mol of CO2 giving 1 mol of CaCO3)

BaO + CO2 --> BaCO3 (1 mol of BaO reacts with 1 mol of CO2 giving 1 mol of BaCO3)

Since CaO & BaO reacts with CO2 in 1:1 ratio, we can tell (x/56.1) mol of CaO reacts with (x/56.1) mol of CO2 and {(2-x)/153.3} mol of BaO reacts with {(2-x)/153.3} mol of CO2.

So, (x/56.1)mol + 1{(2-x)/153.3} mol = 0.01563 mol

Solving for ‘x’ we get, x = 0.2286 g

No. of moles of CaO = x/56.1 = 0.2286/56.1 = 0.0041 mol (n1)

No. of moles of BaO = (2-x)/153.3 = (2-0.2286)/153.3 = 0.1156 mol (n2)

Mole fraction of CaO = n1/(n1+n2) = 0.0041/(0.0041+0.1156) = 0.0343

Mole fraction of BaO = (1 – mole fraction of CaO) = (1- 0.0343) = 0.9657