Question

A mixture of methane and oxygen gases, at a total pressure of 836 mm Hg, contains 4.16 grams of methane and 3.19 grams of oxygen. What is the partial pressure of each gas in the mixture?

P_CH4 = _mm Hg
P_O2 = _ mm Hg

Answer 1

Solution : Formula : Partial pressure mole fraction & Totul pressure d, гаад л . $jax A - P.P= Partial pressure of A Р), 'ХХР), Now, TO Xa=mole fraction of A - Total pressure = 836 mm Hg Pq = Total pressure S of mixture weight of methane (CH W = 4.169 Weight of Oxygen (O2) = 2 3.19g Mole= Weight molecular mass to using we get, Molo mass of methanesi6g (Chu (12+4) mol-mass of oxygen = 32g (02) mole of a 3.190.0997 mole 02 22 ( 32 mole of , 4.16 = 0.26 mole (methane) Chy . To mole fraction = of A mole or A Total moles Then, Total moles = moles of Chu & moles of O₂ 3 0.26 + 0.0997 : 0.3592 mole w fraction of O2: 0.0997 -0.2772 using ③, we get on a mole fraction 0.3547 mole fraction of Chy> 0.26 0.3597 E = 0.7228

Using eq , We get Partial mole fraction of CHu X Total pressure pressure CHy te 0.7 228 X 836 = 604.3 mm Hg. - Partial pressure of we of, mole fraction of O₂ * Total pressure m 0.2772 X 836 231-7 mm Hg Ans) Peru = 604.3 mm Hg Por > 231.7 mm Hg
Please give me better rattings.

Thank you.