Question

A 4.10 −g sample of a mixture of CaO and BaO is placed in a 1.00-L vessel containing CO2 gas at a pressure of 735 torr and a temperature of 26 ∘C. The CO2 reacts with the CaO and BaO, forming CaCO3 and BaCO3. When the reaction is complete, the pressure of the remaining CO2 is 155 torr .Calculate the number of moles of CO2 that have reacted.Calculate the mass percentage of CaO in the mixture.

Answer 1

Answer:

The mass % of CaO is 9.37%%

Explanation:

Step 1: Data given

Mass of the mixture = 4.10 grams

Volume of the vessel = 1.00 L

The vessel contains CO2 with a pressure of 735 torr and temperature = 26 °C

The pressure of the remaining CO2 is 155 torr

Step 2: The balanced equation

CaO/BaO + CO2 → CaCO3 /BaCO3

.

Step 3: Calculate moles CO2 (via ideal gas law)

CO2 is the only gas, this means the drop in the pressure is caused by CO2

Δp = 735 torr - 155 torr = 580 torr

580 torr = 0.763 atm

n = PV/RT = (0.763atm)(1.00L) / (0.08206 L*atm/K*mol)(299) = 0.0391 moles CO2 initial

moles reacted = 0.0311 moles CO2

Step 4: Calculate mass of CaO and BaO

Let's suppose  x = mass of CaO. Then mass of BaO = 4.10-x.

In the balanced equation, we notice the mole ratio of CaO/BaO with CO2 is 1:1. This means the number of moles CaO + moles BaO = Number of moles CO2

moles CaO + moles BaO = moles CO2

x/56.08 + (4.10-x)/153.33 = 0.0311

56.08 (x/56.08 + (4.10-x)/153.33) = (56.08)(0.0311)

x + (0.366)(4.10-x) = 1.744

x + 1.5006 - 0.366x = 1.744

0.634x = 0.2434

x = 0.384g CaO

4.1-x = g BaO = 3.716 grams

%CaO = (0.384 g / 4.10 g) x 100 = 9.366% ≈ 9.37

The mass % of CaO is 9.37%