Question

CaO(s) reacts with SO2 (g) according to the following equation: CaO(s) + SO2 = CaSO3(s). A 8.85 g mixture containing CaO(s) is placed in a 3.0L container. The container is filled with SO2(g) to a pressure of 749 torr, at 20 C. After the reaction has gone to completion, the pressue inside the flask is 390 torr also at 20 C. What is the mass of CaSO3(s) produced? assume that only CaO(s) in the mixture reacts with SO2(g).

Answer 1

Given, the balanced chemical reaction,

CaO(s) + SO₂(g) CaSO₃(s)

Mass of mixture containing CaO(s) = 8.85 g

The volume of the container = 3.0 L

The initial pressure of SO₂ gas in the container = 749 Torr x ( 1atm /760 Torr) = 0.9855 atm

Temperature(T) = 20 ^oC + 273.15 = 293.15 K

Final pressure of SO₂ gas in the container = 390 Torr x ( 1atm / 760 Torr) = 0.5132 atm

Calculating the initial moles of SO₂ gas from the given data,

We know, the ideal gas law equation,

PV = nRT

Here, R = 0.08206 L.atm /mol. K

Rearranging the formula,

n = PV /RT

Substituting the known values,

n = (0.9855 atm x 3.0 L) /(0.08206 L.atm/mol.K x 293.15 K)

n = 0.1229 mol

Now, calculating the number of moles of SO₂ remained after the reaction has gone to completion,

n = PV /RT

Substituting the known values,

n = (0.5132 atm x 3.0 L) /(0.08206 L.atm/mol.K x 293.15 K)

n = 0.064 mol

Now, calculating the number of moles of SO₂ reacted with the CaO(s),

= Initial number of moles of SO₂ - Moles of SO₂ remained after reaction completion

= 0.1229 mol - 0.064 mol

= 0.0589 mol of SO₂ gas

Assuming that only CaO(s) in the mixture reacts with SO₂(g)

Moles of CaO reacts with SO₂ = 0.0589 mol [ since the mole ratio between CaO and SO₂ is 1:1 ]

Now, using the moles of of CaO and the mole ratio from the balanced chemical reaction,calculating the number of moles of

CaSO₃ forming,

= 0.0589 mol CaO x ( 1 mol CaSO₃ / 1 mol of CaO)

= 0.0589 mol of CaSO₃

Converting number of moles to grams,

= 0.0589 mol of CaSO₃ x (120.17 g /1 mol)

= 7.08 g of CaSO₃(s) is produced