The density of the product gas mixture is 5.39 g/L
Explanation:
Given:
Temperature after the reaction = [tex]25^0[/tex]C
Pressure in the container = 1.65 atm
The ideal gas equation PV = nRT
To find: Density of the product gas mixture
Step 1:
Molarity (M) = P/(RT)
[tex]\[M=\frac{P}{R T}\]$M=\frac{1.65 \mathrm{atm}}{\left(0.08206 \frac{L \cdot a t m}{m o l . K}\right)(298.15 K)}$[/tex]
M = 0.0674 mol/L
Step 2:
The product of the given reaction is [tex]SO_3[/tex]. Its molar mass is
1S x 32.066 g/mol = 32.066 g/mol
3O x 16 g/mol = 48 g/mol
Adding both we get 80.066 g/mol
Therefore the density of the product gas mixture is [tex]0.0674 \frac{{mol}}{L} \times \frac{80.066 g}{1 \mathrm{mol}}[/tex]
= 5.39 g/L