How Do I Calculate the Mass Proportion of Oxygen in the Air for This Reaction?

  • Thread starter Thread starter marellasunny
  • Start date Start date
  • Tags Tags
    Mass
Click For Summary
SUMMARY

The discussion focuses on calculating the mass proportion of oxygen in the air for the combustion reaction of a hydrocarbon fuel represented by the formula C_x H_y O_z. The key equation derived is $$\xi_{O_2,air}=(x+\frac{y}{4}-\frac{z}{2}) * \frac{M_{O2}}{M_{air/fuel?}}$$, which determines the moles of O2 required per mole of fuel. The calculation is confirmed as correct, emphasizing the need to use atomic masses to derive the mass proportions accurately. Additionally, the relationship between moles of nitrogen and oxygen in air is highlighted as crucial for determining the overall mass proportions.

PREREQUISITES
  • Understanding of stoichiometry in chemical reactions
  • Familiarity with molecular weights and atomic masses
  • Knowledge of combustion reactions and fuel formulas
  • Basic principles of gas mixtures, particularly air composition
NEXT STEPS
  • Calculate the atomic masses of carbon, hydrogen, oxygen, and nitrogen for accurate mass proportion calculations
  • Learn about the ideal gas law and its application in stoichiometric calculations
  • Explore the concept of molar ratios in combustion reactions
  • Study the composition of air and its impact on combustion efficiency
USEFUL FOR

Chemistry students, chemical engineers, and anyone involved in combustion analysis or fuel efficiency optimization will benefit from this discussion.

marellasunny
Messages
245
Reaction score
3

Homework Statement



Fuel general formula: C_x H_y O_z

How do I calculate the mass proportion of oxygen in the air participating in the following reaction \xi_{O_2,air}?
$$C_x H_y O_z+(x+\frac{y}{4}-\frac{z}{2}).O_2 \rightarrow xCO_2+\frac{y}{2}H_2O$$

Homework Equations



Mass proportion of oxygen in fuel:
$$Oxygen=\frac{M_{oxygen}}{M_{fuel}} .z $$
M-molecular weight
z- stoichiometric coefficient

The Attempt at a Solution

The number of moles in the air would be:(x+\frac{y}{4}-\frac{z}{2})

So,$$ \xi_{O_2,air}=(x+\frac{y}{4}-\frac{z}{2}) * \frac{M_{O2}}{M_{air/fuel?}}
$$

Is this correct? Is the mass proportion calculated above EQUAL to the stoichiometric air requirement?
 
Last edited:
Physics news on Phys.org
marellasunny said:

Homework Statement



Fuel general formula: C_x H_y O_z

How do I calculate the mass proportion of oxygen in the air participating in the following reaction \xi_{O_2,air}?
$$C_x H_y O_z+(x+\frac{y}{4}-\frac{z}{2}).O_2 \rightarrow xCO_2+\frac{y}{2}H_2O$$


Homework Equations



Mass proportion of oxygen in fuel:
$$Oxygen=\frac{M_{oxygen}}{M_{fuel}} .z $$
M-molecular weight
z- stoichiometric coefficient

The Attempt at a Solution




The number of moles in the air would be:(x+\frac{y}{4}-\frac{z}{2})

So,$$ \xi_{O_2,air}=(x+\frac{y}{4}-\frac{z}{2}) * \frac{M_{O2}}{M_{air/fuel?}}
$$

Is this correct? Is the mass proportion calculated above EQUAL to the stoichiometric air requirement?

Correct? It's a correct calculation of moles of O2 consumed per mole of this fuel.
That is just the first step to getting the answer requested as I understand it, which is about masses. You will need to use atomic masses.

To answer this question it is not necessary to use the fact that oxygen is diatomic. It does no harm and you have done it correctly but it is just an unnecessary complication for the purposes of answering the question.
 
Take as a basis one mole of fuel. You already showed that 1 mole of fuel requires (x+y/4-z/2) moles of oxygen. How many moles of nitrogen are there for every mole of oxygen in air? How many moles of N2 are carried along for every mole of fuel? On the basis of 1 mole of fuel, what is the seight of the fuel? What is the weight of the oxygen from the air? What is the weight of the nitrogen from the air? What fraction of the total weight is oxygen from the air?

Chet
 

Similar threads

Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
10K
Replies
19
Views
2K
Replies
3
Views
2K