An \"ideal\" fuel/air mixture in which both the fuel and the oxygen in the air a
ID: 942531 • Letter: A
Question
An "ideal" fuel/air mixture in which both the fuel and the oxygen in the air are completely consumed is called the "stoichiometric" mixture. The stoichiometric mixture is usually specified as the ratio of the mass of air to the mass of a particular fuel. For example, if a fuel (for example, ethanol, C2H;OH) required 9 kg of air to precisely burn 1 kg of the fuel (that is, all the fuel burned and there was no oxygen left in the air afterwards) then the stoichiometric ratio for that fuel would be 9: 1 Calculate the partial pressures of the ethanol and oxygen in the mixture based on the stoichiometric ratio given above at 293.15 K. The volume of the internal combustion engine is considered to be 2.000 L. The mass of the fuel is 400.0 g Mass fraction of oxygen in air is 23.1% Molar mass of ethanol is 46.069 g/mol ethanol atm atm oxygen Submit Answer Try Another Version 4 item attempts remainingExplanation / Answer
3Mg + N2 ---> Mg3N2
2Mg + O2 ----> 2MgO
Moles of Mg = mass/molar mass = 3.55/24 = 0.148
Now, let x g of MgO is formed and (5.09-x) g of Mg3N2 be formed
Thus, moles of MgO = x/40 and moles of Mg3N2 = (5.09-x)/100
Now, total moles of Mg reacting = 3*moles of Mg3N2 formed + moles of MgO formed
or, 0.148 = (3)*[(5.09-x)/100] + (x/40)
or, x = 0.94 g = mass of MgO
mass of Mg3N2 formed = 4.15 g
2) mass of air required = 9*400 = 3600 g
mass of O2 = 0.231*3600 = 831.6 g
moles of ethanol = mass/molar mass = 400/46.069 = 8.683
moles of O2 = mass/molar mass = 831.6/32 = 25.9875
Total moles of gas mixture = 34.6705
mole fraction of O2 = moles of O2/moles of gas mixture = 0.75
mole fraction of ethanol = 1 - mole fraction of O2 = 0.25
Now, total pressure of gas mixture, P = n*R*T/V = 34.6705*0.082*293.15/2 = 417.23 atm
Thus, partial pressure of ethanol = mole fraction*total pressure = 104.304 atm
partial pressure of O2 = mole fraction*total pressure = 312.914 atm
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