Answer:
The mass % of CaO is 9.37%%
Explanation:
<u>Step 1: </u>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
<u>Step 2:</u> The balanced equation
CaO/BaO + CO2 → CaCO3 /BaCO3
.
<u>Step 3</u>: 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%
