SOLVING CHEMISTRY QUESTION GRADE 11 FSC PRE MED

Question

Three gases (8.00g of methane, CH4, 18.0g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0L container. At 23.0°C, the total pressure in the container was measured to be 4.43 atm. Calculate the partial pressure of each gas in the container.

Answer 1

8g methane = 8/16=0.50 mol

18g ethane = 18/30=0.6 mol

unknown propane, molar mass = 44 g/mol

total pressure = 4.43

Ptotal=Pmethane+Pethane+Ppropane

PV=nRT

P=atm

V=Liters

n=mol

R=0.0821 Latm/molK

T=Kelvin

PmethaneV=nRT

P(10.0)=0.50(0.0821)(296)

=(0.50*0.0821*296)/10

Pmethane=1.21508=1.22 atm

PethaneV=nRT

P(10.0)=0.60(0.0821)(296)

=(0.60*0.0821*296)/10

Pethane=1.458096=1.5atm

Ptotal=Pethane+Pmethane+Ppropane

4.43=1.5+1.22+Ppropane

Ppropane=4.43-1.5-1.22

Ppropane=1.71atm