The mole fraction of oxygen molecules in dry air is 0.2095, whatvolume of dry ai

Question

The mole fraction of oxygen molecules in dry air is 0.2095, whatvolume of dry air at 1.00 atm, and is required for burning 1.00 Lof hexane (C6H12, density = 0.660 g/mL)completey, yielding carbon dioxide and water? A balanced equationfor this process is given by: 2C6H14(l) + 19O2(g) -->12CO2(g) + 14H2(l) Well, I asked this before, but I figured that one of theanswers wasn't correct, this is a multiple choice question, BEWARE,THERE ARE FALSE ANSWERS A. 187L B. 712L C. 1780 L D. 894 L E. 8490 L This is what I did that isn't correct, or which could becorrect, not sure: The Question Says: P = 1atm V = 1L d = 0.660 g/mol molar mass = 86.06 g (Hexane) This gives me how much Hexane that I would need: 0.660/1000ml = 660g/86.06 mol/g = 7.66 mol Moles of O2 required are: (7.67 mol hexane) (19 mol O2/2 mol hexane)[Ratio] = 72.865moles of air However, the question says that the mole fraction of dry airare 0.2095 Therefore using the ideal gas equation, pv=nrt (1)(v)=(72.865/0.2095)(0.0821)(25+273.15) v=8513.59L Which isn't on the multiple choice, and by the way, if you doit in 20 significant figures, then you would get about 8700.00 L,which isn't correct. So, what's the right way to do it? 2C6H14(l) + 19O2(g) -->12CO2(g) + 14H2(l) Well, I asked this before, but I figured that one of theanswers wasn't correct, this is a multiple choice question, BEWARE,THERE ARE FALSE ANSWERS A. 187L B. 712L C. 1780 L D. 894 L E. 8490 L This is what I did that isn't correct, or which could becorrect, not sure: The Question Says: P = 1atm V = 1L d = 0.660 g/mol molar mass = 86.06 g (Hexane) This gives me how much Hexane that I would need: 0.660/1000ml = 660g/86.06 mol/g = 7.66 mol Moles of O2 required are: (7.67 mol hexane) (19 mol O2/2 mol hexane)[Ratio] = 72.865moles of air However, the question says that the mole fraction of dry airare 0.2095 Therefore using the ideal gas equation, pv=nrt (1)(v)=(72.865/0.2095)(0.0821)(25+273.15) v=8513.59L Which isn't on the multiple choice, and by the way, if you doit in 20 significant figures, then you would get about 8700.00 L,which isn't correct. So, what's the right way to do it?

Explanation / Answer

What you've done is right. I think its just a matter ofrounding off and the answer is (e) 8490 L . Mass of hexane = 1.00 L * ( 1000 mL / 1 L) * (0.660 g / 1mL)                         =660 g . Moles of hexane = 660 g / 86.17 g/mol = 7.659 moles . Moles of oxygen = 7.659 moles hexane * ( 19 moles O2 / 2 molesHexane)                          = 72.7 moles O2 . Mole fraction of O2 = 0.2095 Mole fraction of O2 = moles of O2 / Moles of air 0.2095 = 72.7 moles / moles of air Moles of air = 347 moles . Volume of air , V = n R T / P                            =347 moles * 0.0821 L-atm / (mol.K) * 298 K / 1.00 atm                            ˜8490 L

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