Question

Constants | Periodic Table
A 3.80 −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 .

Part A

Calculate the number of moles of CO2 that have reacted.

n =    mol  

Part B

Calculate the mass percentage of CaO in the mixture.
Express your answer using two significant figures.

%

Answer 1

Using the gas equation:
PV = nRT
0.97 * 1 = n* 0.082057 * 299
n =0.039 mol CO2

Calculate mol of CO2 present at end:
PV = nRT
0.204 *1 =n*0.082057 * 298
n = 0.00834 mol

mol of CO2 reacted = 0.039-0.00834 = 0.0307 mol CO2 consumed

b)
Mass of CO2 consumed:
We know that the molar mass CO2 = 44.009g/mol
0.0307mol = 44.009*0.0307 = 1.35
Hence the mass of final mixed carbonates = 5.15g

Let the mass of CaO = X
then mass of BaO = (3.80-X)
Molar mass CaCO3 = 100.0875 g/mol
Molar mass CaO = 56.0778 g/mol

Molar mass BaCO3 = 197.3368 g/mol
Molar mass BaO = 153.3271 g/mol

Equation:
(100.0875/56.0778)*X + (197.3368/153.3271)*(3.80-X) = 5.15
1.78X + 1.28(3.80-X) = 5.15
1.78X +4.864 - 1.28X = 5.15
0.5X = 0.286
X = 0.572 g

Mass of CaO = 0.572g

%CaO in sample = 0.506/3.80*100

= 13.31 %