Mean molar mass after combustion in a combustion engine

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SUMMARY

The discussion focuses on calculating the mean molar mass of a combustion reaction involving butane (C4H10) in a combustion engine. The reaction includes nitrogen (N2) and oxygen (O2) as reactants, with the combustion products being water (H2O) and carbon dioxide (CO2). The user, Christopher, has attempted to balance the reaction and calculate the mean molar mass using the masses of air and fuel provided but has not achieved the correct answer. The key molar masses involved are MH2O = 18 g/mol, MCO2 = 44 g/mol, MN2 = 28 g/mol, and MO2 = 32 g/mol.

PREREQUISITES
  • Understanding of chemical reaction balancing
  • Knowledge of molar mass calculations
  • Familiarity with combustion reactions
  • Basic skills in stoichiometry
NEXT STEPS
  • Learn how to balance combustion reactions involving hydrocarbons
  • Study the concept of mean molar mass in chemical reactions
  • Explore stoichiometric calculations in combustion processes
  • Review the properties and calculations of gases under standard conditions
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Chemistry students, chemical engineers, and anyone involved in combustion analysis or fuel efficiency studies will benefit from this discussion.

ChristopherJ
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Homework Statement


The problem is to calculate the mean molar mass of the reaktion below, which takes place in a combustion engine.

C4H10 + z(0.79N2 + 0.21O2) → αH2O + βCO2 + γN2

where the molar masses for the different molecules/atoms are:
MH2O = 18, MCO2 = 44, MN2 = 28, MO2 = 32, MC = 12, MH = 1.

The mass of the air in the cylinder before combustion is 2.39905 grams. The mass of the fuel in the cylinder is 45.0 milligrams.

Homework Equations


See above


The Attempt at a Solution


I have tried to balance the equation above by calculating the amount of molecules of oxygen and nitrogen in the cylinder and the amount of molecules in the fuel to get α, β and γ. And then use these values to balance the reaction equation above. The mean molar mass can be calculated by dividing the right hand side by the amount of molecules at hand.

But, I don't get the right answer by doing this. Can anyone help me with this?

I am very grateful for answers!

Best regards
Christopher
 
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