The mole fraction of oxygen molecules in dryair is 0.2095, what volume of dry ai

Question

   The mole fraction of oxygen molecules in dryair is 0.2095, what volume of dry air at 1.00 atm, and is requiredfor burning 1.00 L of hexane (C6H12, density= 0.660 g/mL) completey, yielding carbon dioxide and water? Abalanced equation for this process is given by: 2C6H14(l) + 19O2(g) -->12CO2(g) + 14H2(l)    The mole fraction of oxygen molecules in dryair is 0.2095, what volume of dry air at 1.00 atm, and is requiredfor burning 1.00 L of hexane (C6H12, density= 0.660 g/mL) completey, yielding carbon dioxide and water? Abalanced equation for this process is given by: 2C6H14(l) + 19O2(g) -->12CO2(g) + 14H2(l)

Explanation / Answer

Mass of hexane = 1.0L (0.660g/mL) = 0.660x10^3g moles of hexane = mass / molar mass = 0.660 x10^3g /(86.18g/mol) = 7.67mol Moles of O2 required = 7.67mol hexane *(19 molO2 / 2mol hexane) = 72.865molO2 Moles of O2 required = molefraction ofO2 * moles of air    Moles of O2 required = molefraction ofO2 * (PV /RT) So, volume of air,V =   Moles ofO2 required * RT / (molefraction ofO2 * P)   In the data:      molefractionof O2 , P (of air) given but T of air not given soasuming T = 250C = 298K                   Volume of air needed wolud be ,V=   (72.865mol) (0.0821Latm mol^-1 K^-1)(298K)/ (0.2095 *1atm)   Moles of O2 required = molefraction ofO2 * (PV /RT) So, volume of air,V =   Moles ofO2 required * RT / (molefraction ofO2 * P)   In the data:      molefractionof O2 , P (of air) given but T of air not given soasuming T = 250C = 298K                   Volume of air needed wolud be ,V=   (72.865mol) (0.0821Latm mol^-1 K^-1)(298K)/ (0.2095 *1atm)  

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