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

Given P = 735 torr = 735 / 760 atm = 0.967 ( 1atm = 760 torr)

V = 1L ,   T = 26C = 26+273 = 299 K

we use PV = nRT to get moles of CO2 = n

0.967 atm x 1L = n x 0.08206 liter atm/molK x 299 K

n = 0.0394 = CO2 moles present initially

after compl;eteion of reacton P = 155 torr = 155 /760 = 0.204 atm

0.204 atm x 1L = n x 0.08206 liter atm /molK x 299 K

n = 0.008312 = CO2 moles left

CO moles consumed = 0.0394 - 0.008312 = 0.031

Let CaO moles = a

CaO mass= moles x molar mass of Cao = a x 56 g/mol = 56 a

BaO moles = b , then BaO mass = moles x molar mass of BaO = b x 153.3 = 153.3 b

mass of CaO + BaO mass = 4.1

56a + 153.3 b= 4.1 ......(1)

CO2 moles reacted with CaO = a   ( since CaO and CO2 react in 1:1 ratio)

CO2 moles reacted with BaO = b

now moles of CO2 reacted   a+b = 0.031 ..(2)

by (1) ( 2) we get a = 0.0067

CaO mass = 0.0067 mol x 56 g/mol = 0.3752 g

CaO % = ( 100 x CaO mass) / smple mass

         = ( 100 x 0.3752g / 4.1 g)

         = 9.15 %