If 48.0 g of O2 and 4.4 g of CO2 are placed in a 10.0 L container at 21 C, what is the pressure of this mixtur [Solved]

P (atm) = 1.6 * 0.082057 * 295/10 (mol * L atm K-1 mol-1 * K / L) = 3.87 atm. which sounds about proper as 1 mol occupies 22.4 L at stp.

Mol wt O2 = 32 g so 48 g = 1.5 moles
Mol wt CO2 = 44 g so 4.4 g = 0.1 moles so whole 1.6 moles. PV = nRT. P = nRT/V.
V = 10 L; T = 273 21 = 295 K ; n = 1.6 moles; R = 0.082057 LatmK1mol1

P (atm) = 1.6 * 0.082057 * 295/10 (mol * L atm K-1 mol-1 * K / L) = 3.87 atm. which sounds about proper as 1 mol occupies 22.4 L at stp.

This post is last updated on hrtanswers.com at Date : 1st of September – 2022