Extra work # 2 1) Identify the hydrocarbon liquid which is considered as a membe
ID: 590817 • Letter: E
Question
Extra work # 2 1) Identify the hydrocarbon liquid which is considered as a member o the cycloalkane family. The vapor of this liquid has a density of 2.64 g /L measured at 101.325 Kpa and 49. 5 points 2) Consider the combustion reaction between 25.0 mL of liquid methanol (CH3OH, density = 0.85 g / mL) and 12.5 L of oxygen gas measured at STP. The product of reaction are CO, (gas) and H2 (gas). Calculate the volume of liquid H,0 formed if the reaction goes to completion and you condense the water vapor. points 3) In a more recent developed process, ethanol can be prepared by reacting ethylene (found in petroleum) with water. If 100.8 L of ethylene measured @ STP reacts with water vapor. It is found that only 186.3 g of ethanol is produced, what is the percent yield of the reaction? 5 pointsExplanation / Answer
from gas law, PV= nRT
PV= (mass/molar mass)*RT, P* molar mass= (mass/volume)*RT, P* molar mass= density*RT
molar mass= density*RT/Pressure
P= presssure taken in atm= 101.325 Kpa, in terms of atm, P= 101.325/101.3=1atm,
T= 49 deg.c= 49+273= 322K, R= 0.0821 L.atm/mole.K , density= 2.648 g/l
hence molar mass= 2,648*0.0821*322/1=70
cyclohexane have general formula of (CH2)n, CH2 is empirical formula and n= no of repeating empirical formulas
molar mass of CH2= 12+4= 14
n*14= molar mass of cyclocompound= 70
n= 5
hence molecular formula is C5H10, cyclopentane.
2. mass of CH3OH= volume*density = 25ml*0.85 g/ml =21.25gm, moles of CH3OH= mass/molar mass= 21.25/32 =0.664 moles. 1 mole of any gas at STP occupies 22.4 L, 12.5 L of Oxygen gas will have 12.5/22.4 moles = 0.56 moles
the reaction of combustion of CH3OH with oxygen is CH3OH+ 1.5O2------->CO2+ 2H2O
theoreitical molar ratio of CH3OH: O2= 1:1.5 actual ratio given = 0.664:0.56 = 1:0.56/0.664
Oxygen is limiting and excess is CH3OH. All the O2 reacts.
As per the reaction 1.5 mole of O2 gives 2 moles water. 0.56 moles of oxygen gas will have 2*0.56/1.5 =0.75 moles of oxygen. at STP volume of oxygen formed = 0.75*22.4 L16.8 L
3. the reaction is C2H4(ethylene)+H2O------->C2H5OH (ethanol)
22.4 L are there in 1mole of any gas, 100.8 L are there in 100.8/22.4 moles of ethylene =4.5 moles of ethylene
moles of ethanol formed as per the reaction = 4,5 moles, mass of ethanol formed= moles* molar mass = 4.5*46=207 gm. this is the theoretical amount. % yield =100* actual amount formed/ theoretical amount= 100*186.3/207=90%
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