What is the molar mass of a gas if 372.0 mL has a mass of 0.800 g at 99.8°C and 106.6 kPa?
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Answer:
Apply the Ideal Gas Law : --------------------------------------… ( P ) ( V ) = ( m ) ( R ) ( T ) / ( M ) M = ( m ) ( R ) ( T ) / ( P ) ( V ) M = ( 0.800 ) ( 0.008314 ) ( 99.8 + 273.2 )/ ( 106.6 ) ( 0.372 /1000 ) M = 62.56 g / gmol <---------------------------------------… **************************************… C2H6 + ( 7/2 ) O2 ----> 2 CO2 + 3 H2O .. { Mole basis ] 30 C2H6 + 112 O2 ---> 88 CO2 + 54 H2O n sub CO2 = ( 1.05 C2H6 / 30 ) ) ( 2 CO2 / C2H6 ) n sub CO2 = 0.070 kgmol = 70.00 gmol <-------------------------- V sub CO2 = ( n sub CO2 ) ( R ) ( T ) / ( P ) V sub CO2 = ( 70 ) ( 0.008314 ) ( 287.2 ) / ( 745 / 760 ) ( 101.325 ) V sub CO2 = ( 70 ) ( 0.008314 ) ( 287.3 ) / ( 99.33 ) V sub CO2 = 1.68 L <----------------------------------
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Other answers
molar mass = mRT/PV = (8x10^-4 x 8.314 x (273+99.8))/106.88x10^3 x 0.372 =6.23 10^-6 kg 26. 36 gm C2H6 gives 176 gm CO2 1 ,,,,,,,,,,,,,,,,,,,,,,,,,,,176/36 =4.89 gm CO2 1050 gm 4.899 g x 1050 gm =5133,33 gm CO2 CONVERT THIS gm> moles of co2 5133.33/44 =116.67 moles Now PV=nRT you ahve P=745 mm of Hg =0.98 Pascal V= ? n=116.67 R=8.314 T=24 =(273+24)=297 k so V=nRT/P = 116.67 X 8.314X 296 / 0.98 =292977.89 L
molar mass = mRT/PV = (8x10^-4 x 8.314 x (273+99.8))/106.88x10^3 x 0.372 =6.23 10^-6 kg 26. 36 gm C2H6 gives 176 gm CO2 1 ,,,,,,,,,,,,,,,,,,,,,,,,,,,176/36 =4.89 gm CO2 1050 gm 4.899 g x 1050 gm =5133,33 gm CO2 CONVERT THIS gm> moles of co2 5133.33/44 =116.67 moles Now PV=nRT you ahve P=745 mm of Hg =0.98 Pascal V= ? n=116.67 R=8.314 T=24 =(273+24)=297 k so V=nRT/P = 116.67 X 8.314X 296 / 0.98 =292977.89 L
Himalayan Sheperd
Apply the Ideal Gas Law : --------------------------------------... ( P ) ( V ) = ( m ) ( R ) ( T ) / ( M ) M = ( m ) ( R ) ( T ) / ( P ) ( V ) M = ( 0.800 ) ( 0.008314 ) ( 99.8 + 273.2 )/ ( 106.6 ) ( 0.372 /1000 ) M = 62.56 g / gmol <---------------------------------------... **************************************... C2H6 + ( 7/2 ) O2 ----> 2 CO2 + 3 H2O .. { Mole basis ] 30 C2H6 + 112 O2 ---> 88 CO2 + 54 H2O n sub CO2 = ( 1.05 C2H6 / 30 ) ) ( 2 CO2 / C2H6 ) n sub CO2 = 0.070 kgmol = 70.00 gmol <-------------------------- V sub CO2 = ( n sub CO2 ) ( R ) ( T ) / ( P ) V sub CO2 = ( 70 ) ( 0.008314 ) ( 287.2 ) / ( 745 / 760 ) ( 101.325 ) V sub CO2 = ( 70 ) ( 0.008314 ) ( 287.3 ) / ( 99.33 ) V sub CO2 = 1.68 L <----------------------------------
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