Part A Three gases (8 00 g of methane, CH. 180 g of ethane, CHs, and an urknown

Question

Part A Three gases (8 00 g of methane, CH. 180 g of ethane, CHs, and an urknown amount of propane, CHs) were added to the same 10 0-L t hae0L. container. At 23 0the total pressure in the container is 5 50 atm calculate the partial pressure of each gas i then propane. View Available Hint s) Espres the pressure valoes numericaly in atmospheres, spardbcas. Enter he partisl pressure of methane first, hen ethane, Part B Agase us mo ture of O and N2 contains 40 8 % n rogen by mass what is the partial pressure ofony e inthe mixture 'the total pressure is 645 mmig? Express you answer numerically in millimeters of mercury View Available Hints) F11 F12 6 7 8 8 9

Explanation / Answer

mass of CH4=8.00 grams

molar mass of CH4= 16.0grams

number of moles of CH4= 8.00/16.0= 0.5 moles

mass of C2H6 = 18.0 grams

molar mass of C2H6= 28 gram/mole

number of moles of C2H6 = 18.0/28.0 =0.64 moles

P=5.50 atm

V=10.0L

T=23C= 23+273= 296K

R= 0.0821 L-atm/mole-K

PV= nRT

n= PV/Rt

n= 5.50x10.0/0.0821x296= 2.26 moles

number of moles of C3H8 = 2.26 moles

total number of moles = 0.5+0.64+2.26=3.4 moles

molefraction of CH4= number of moles/total number o f moles

molefraction of CH4= 0.5/3.4=0.147

molefraction of C2H6= 0.64/3.4=0.188

molefraction of C3H8= 2.26/3.4=0.665

partial pressure of CH4= total pressure x molfraction of CH4

partial pressure of Ch4 = 5.50x0.147 = 0.8085 atm

partial pressure of C2H6 = 5.50x0.188 = 1.034 atm

partial pressure of C3H8= 5,50x0.665= 3.66 atm

part-b

% by mass of N2= 40.8%

%by mass of O2= 100-40.8= 59.2%

total pressure =645 mm

partial pressure of O2 = 645x59.2/100 = 381.84 mm

partail pressure of O2= 381.84mm

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