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

• marellasunny
In summary: Final ThoughtsIn summary, the mass proportion of oxygen in air participating in the following reaction is calculated as (x+y/4-z/2)*(frac{M_{O2}}{M_{air/fuel?}}).
marellasunny

## 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:
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

## What is mass proportion in reactant?

Mass proportion in reactant refers to the relative amounts of each reactant involved in a chemical reaction. It is expressed as the ratio of the mass of one reactant to the mass of another reactant.

## Why is mass proportion in reactant important?

Understanding the mass proportion in reactant is important because it helps predict the amount of products that will be formed in a chemical reaction. It also allows for the calculation of the yield of a reaction and helps determine the limiting reactant.

## How is mass proportion in reactant calculated?

To calculate mass proportion in reactant, the molar masses of each reactant are determined and then converted to moles. The moles of each reactant are then compared to determine their ratio, which represents the mass proportion in reactant.

## Can mass proportion in reactant change during a reaction?

Yes, the mass proportion in reactant can change during a reaction as reactants are consumed and products are formed. This is because the relative amounts of each reactant are constantly changing.

## How does mass proportion in reactant affect the rate of a reaction?

The mass proportion in reactant can affect the rate of a reaction because it determines the concentration of the reactants. A higher concentration of reactants usually leads to a faster reaction rate, while a lower concentration may slow down the reaction.

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