Question

3- (15 pts) A mixture of gases contains 4.96 grams of CH4, 7.5 grams of C2He, and 12.76 grams of CaHs. The total pressure is 1.50 atm. Calculate the partial pressure of each gas.

Answer 1

Answerg- mixture of gases CHu & 2" 1. & C3Hg given mass of CH4 = 40969 molar mass of CH = 16.04 8l mol . 4.96 no" of moles (richa) = 16-04 Mch. - 0-309 moles given mass of Calho = 7.59 molar mass of Ca Hg = 30.01g/mol no" ot moles incite=70.07 Moto=0-2494 of fonoles) given mass of C3 Hg = 12-16 molar mass of Is Hg = 4401 hoot moles of C3Hg = n = 12.76 = 0-289 • ? Hg uyol mole

Total Toso atm CHy Hi Total nity Ncht hatt mange X 1050 0-309 0-309 +0.2494 +0.289 - 00309 TH 0-8474 ** 1 Pet = 0.5469 atmi noalto Caho - x Protal Metathathoth late 0.2494 x 1.50 =07357 atm 0,8474 PC2H6 Pratte=0.uulu atm 1

P =Xcata Protal "Ca Hg Prehe Mests x Protal ; netut health cate Puc 0.289 x1-50 Pigits-00 0.8414 Peatte=0.511 atm