Respuesta :
Explanation:
The given reaction is as follows.
[tex]SO_{2} + O_{2} \rightarrow SO_{3}[/tex]
Now, balancing the given equation by putting appropriate coefficients.
[tex]2SO_{2} + O_{2} \rightarrow 2SO_{3}[/tex]
It is given that equimolar [tex]SO_{2}[/tex] and [tex]O_{2}[/tex]. Hence,
Therefore, during completion of the reaction,
[tex]SO_{2} + O_{2} \rightarrow SO_{3} + \frac{1}{2}O_{2}[/tex]
Temperature (T) = [tex]25^{o}C[/tex] = (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 [tex]O_{2}[/tex] and [tex]SO_{3}[/tex] remains in the mixture. Therefore, molar mass of the mixture is equal to the sum of molar mass of
Total molar mass = [tex]O_{2} + SO_{3}[/tex]
= (32 + 80) g/mol
= 112 g/mol
Hence, according to the formula we will calculate the density as follows.
Density = [tex]\frac{P \times \text{molar mass}}{R \times T}[/tex]
= [tex]\frac{1.25 atm \times 112 g/mol}{0.0820 Latm/mol K \times 298 K}[/tex]
= 5.73 g/L
Thus, we can conclude that density of the product gas mixture is 5.73 g/L.
Answer:
d = 3.27g/L
Explanation:
[tex]2SO_{2} + O_2 => 2SO_3[/tex]
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
[tex]X_O_2 = \frac{0.5 mol}{1.5 mol} = \frac{1}{3}\\\\X_S_O_3 = \frac{1 mol}{1.5 mol} = \frac{2}{3}\\[/tex]
[tex]\frac{n}{V} = \frac{P}{RT} = 0.0511 mol / L\\\\d = \frac{n}{RT} * X_O_2 * MM_O_2 + X_S_O_3 * MM_S_O_3\\\\d = (0.0511 mol/L) * (\frac{1}{3} mol * 32.0 g/mol + \frac{2}{3} * 80.06 g/mol)\\\\d = 3.27 g/L[/tex]
