Question

You start with an equimolar mixture of the gases SO₂ and O₂ in a container. The SO₂ and O₂ react to completion to form SO₃. If the temperature after the reaction is 25°C and the pressure in the container is 1.25 atm, what is the density of the product gas mixture? Assume ideal behavior.

Answer 1

Step-by-step explanation:

The given reaction is as follows.

`SO_(2) + O_(2) \rightarrow SO_(3)`

Now, balancing the given equation by putting appropriate coefficients.

`2SO_(2) + O_(2) \rightarrow 2SO_(3)`

It is given that equimolar
`SO_(2)` and
`O_(2)`. Hence,

Therefore, during completion of the reaction,

`SO_(2) + O_(2) \rightarrow SO_(3) + (1)/(2)O_(2)`

Temperature (T) =
`25^(o)C` = (25 + 273) K = 298 K

Pressure (P) = 1.75 atm

R (gas constant) = 0.0820 L atm/mol K

As on completion of reaction there is
`O_(2)` and
`SO_(3)` remains in the mixture. Therefore, molar mass of the mixture is equal to the sum of molar mass of

Total molar mass =
`O_(2) + SO_(3)`

= (32 + 80) g/mol

= 112 g/mol

Hence, according to the formula we will calculate the density as follows.

Density =
`\frac{P * \text{molar mass}}{R * T}`

=
`(1.25 atm * 112 g/mol)/(0.0820 Latm/mol K * 298 K)`

= 5.73 g/L

Thus, we can conclude that density of the product gas mixture is 5.73 g/L.

Answer 2

d = 3.27g/L

Step-by-step explanation:

`2SO_(2) + O_2 => 2SO_3`

ICF table

2SO_{2} + O_2 => 2SO_3

initial 1 mol 1 mol 0 mol

change -1 mol 0.5 mol 1 mol

final 0 mol 0.5 mol 1 mol

calculate the mole fractions

`X_O_2 = (0.5 mol)/(1.5 mol) = (1)/(3)\\\\X_S_O_3 = (1 mol)/(1.5 mol) = (2)/(3)\\`

`(n)/(V) = (P)/(RT) = 0.0511 mol / L\\\\d = (n)/(RT) * X_O_2 * MM_O_2 + X_S_O_3 * MM_S_O_3\\\\d = (0.0511 mol/L) * ((1)/(3) mol * 32.0 g/mol + (2)/(3) * 80.06 g/mol)\\\\d = 3.27 g/L`