Question

PLEASEEE HELPPP ASAPP

3. (15 pts) A mixture of gases contains 4.96 grams of CH., 7.5 grams of CzHs, and 12.76 grams of CsHe. The total pressure is 1.50 atm. Calculate the partial pressure of each gas. A 10 pts

Answer 1

The complete step wise calculation of the partial pressures are given as below:-

③ It is given that, Total pressure (P+) = 1.50 atm. mаgg of Chч – Ч:36 Э molecular weight of CH4 = 16 g/mol Number of moles af CH4 = n1 = 4.969 M = 0.31 moles) = 0.31 moles. 16 g/mol

mass of Cat 6 = 7.5g molecular weight of Cal16= 30 g/mol Number of moles al Cat 6 = n2 = 7.5 g on the = 0.25 moles - © = 0.25 moles 30g/mol

mass of C3 H8 = 12.769 molecular weight af C3H8 = 44.1 gimol Number of moles af C3H8 = n3 = 12.76g -=0.29moles 44.1 g/mol or M3 = 0.29 moles ③ So, the total number of moles Mrl= nithe th3 MT= (0.31 +0.25 +0.29) moles = 0.85 moles In = 0.85 moles 07

So, the mole fraction of CHy will be: x = n = (0.31 moles = 0.37 0.85 moles or X, = 0.371 - 3 NT mole fraction of Call 6 will be:- t2 = (mm) = (0-25 moles ) = 0.294 or [X2 = 0.294 - 0

mole fraction of C3H8 will be:- X3 (0.29 moles = 0.34 0.85 moles) an | 13 = 0.34 7 - I So, the partial pressure of CHy will be :- P = X, Pr = (0.37) (1.50 atm) = 0.555 atm or t = 0.555 atm=partial pressure af Chu

Partial pressure of Cah6 will be: Pq = X2 PT = (0,294) × (1.50 atm)=0 uul atm [pe = 0.441 atm = partial pressure of C2H6. partial pressure af C3H8 will be: - b3 = X3 PT = (0.34) (1.50 atm) = 0.51 atm. on 6 = 0.51 atm = Partial pressure af C3H8.