Help with balance of matter with chemical reaction

In summary: The total mass of solid leaving is 76.92 gm, and, since the mass of inserts is 10 gm, the total mass of FeS2 plus Fe2O3 leaving the container is 66.92. Let x = mass of FeS2 exiting and let (66.92-x) = mass of Fe2O3 exiting. In terms of x, what is the number of moles of FeS2 exiting, and what is the number of moles of Fe2O3 exiting? Based on this, in terms of x, what is the number of moles of Fe atoms exiting? But how you obtain the 76,92 g? $$0.
  • #1
Dionisio Mendoza
9
0
The question arises in which procedure to follow, a balance by element or by compound?
The excersise is this:
A mineral of a process contains contains 90% FeS2 and 10% inerts by weight. This materialit is calcined with 25% excess air, according to the reaction:

FeS2 + 02 ► Fe203 + S02

The solid mixture leaves the calciner with 13% inerts by weight and the rest FeS2 and Fe203

a) Calculate the weight of solids leaving the calciner.

b) Determine the% conversion based on FeS2 •

c) Calculate the Orsat analysis of the gases produced.

So far i only be able to averigue the O2 and N2 wheights.
 

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  • #2
Your handwritten solution is unreadable. Please describe what you did, including your reasoning for each step. The only thing I got out of your writeup is that you started by choosing a basis of 100 g for the FeS2. This was a good start.
 
  • #3
What I did first was to make the flow diagram, then calculate the theoretical moles of oxygen with the calculation base of 100g 90% FeS2 in the feed, obtaining
theoretical oxygen = 90g * 11/4 = 2.06
Then calculate the excess oxygen = theoretical oxygen * excess airflow = 2.06 * .25 = .515moles
After the total oxygen = 2.06 + .515 = 2.575moles
with a rule of three I got that the N2 would be 9.68 mole.
Multiply each one of those values by their molecular weight and the result that I obtained added them and got 581.02 grams, which would be total weight of air
 
  • #4
I confirm your moles of O2 and N2. But I get the total number of grams of air as 355.4 gm.

How did you calculate the number of grams of solids leaving the container? I get 76.92 gm. What are the number of moles of Fe atom leaving the container in the solids?
 
  • #5
[QUOTE = "Chestermiller, post: 6134226, miembro: 345636"] Confirmo sus moles de O2 y N2. Pero obtengo el número total de gramos de aire como 355.4 gm.

¿Cómo calculaste la cantidad de gramos de sólidos que salen del contenedor? Tengo 76,92 gm. ¿Cuál es el número de moles de átomo de Fe que sale del contenedor en los sólidos? [/ QUOTE]

Multiplica el número de moles de oxígeno por su peso molecular, como el N2.
2.575 mm * 32 g / mol = 309.98 g
9.68 mm * 28 g / mol = 271.04 g
Y sumando ambos obtuve el aire total.El número de lunares que salen de Fe no lo sé.
 
  • #6
Dionisio Mendoza said:
[QUOTE = "Chestermiller, post: 6134226, miembro: 345636"] Confirmo sus moles de O2 y N2. Pero obtengo el número total de gramos de aire como 355.4 gm.

¿Cómo calculaste la cantidad de gramos de sólidos que salen del contenedor? Tengo 76,92 gm. ¿Cuál es el número de moles de átomo de Fe que sale del contenedor en los sólidos? [/ QUOTE]

Multiplica el número de moles de oxígeno por su peso molecular, como el N2.
2.575 mm * 32 g / mol = 309.98 g

2.575 X 32 = 82.4
 
  • #7
The number of moles of Fe atoms leaving the container in the solids is the same the number of moles of FeS2 entering. (There is no Fe in the gas).
 
  • #8
[QUOTE = "Chestermiller, post: 6134230, miembro: 345636"] 2.575 X 32 = 82.4 [/ QUOTE]
O god I am wrong its 353.44g of total air
 
  • #9
and the moles of the entry if they are .75?
 
  • #10
Dionisio Mendoza said:
and the moles of the entry if they are .75?
Correct. So the total mass of solid leaving is 76.92 gm, and, since the mass of inserts is 10 gm, the total mass of FeS2 plus Fe2O3 leaving the container is 66.92. Let x = mass of FeS2 exiting and let (66.92-x) = mass of Fe2O3 exiting. In terms of x, what is the number of moles of FeS2 exiting, and what is the number of moles of Fe2O3 exiting? Based on this, in terms of x, what is the number of moles of Fe atoms exiting?
 
  • #11
but how you obtain the 76,92 g
 
  • #12
Dionisio Mendoza said:
but how you obtain the 76,92 g
$$0.13 y = 10$$
 
  • #13
Chestermiller said:
$$0.13 y = 10$$
Oh rigth. but in the other I do not follow you
 
  • #14
If the total solids in the exit is 76.92 gm, and the mass of inerts in the exit stream is 10 gm, what is the combined mass of FeS2 and Fe2O3 in the exit stream?
 

1. What is the definition of balance of matter in a chemical reaction?

The balance of matter in a chemical reaction refers to the principle of conservation of mass, which states that matter cannot be created or destroyed, only transformed. This means that the total mass of the reactants must be equal to the total mass of the products.

2. Why is it important to balance the matter in a chemical reaction?

Balancing the matter in a chemical reaction is important because it ensures that the reaction follows the law of conservation of mass. This also allows for accurate calculations of reactant and product quantities, which is crucial for understanding the reaction and predicting its outcomes.

3. How do you balance the matter in a chemical reaction?

To balance the matter in a chemical reaction, you need to adjust the coefficients (numbers in front of the reactants and products) to make sure that the number of atoms of each element is the same on both sides of the reaction. This can be done by using the method of trial and error or algebraic equations.

4. Can a chemical reaction be balanced without changing the chemical formula?

Yes, a chemical reaction can be balanced without changing the chemical formula. This means that the same types and numbers of atoms must be present on both sides of the reaction. However, the coefficients (numbers in front of the formulas) may need to be adjusted to achieve this balance.

5. What are some common mistakes when balancing the matter in a chemical reaction?

Some common mistakes when balancing the matter in a chemical reaction include forgetting to balance the atoms of each element, changing the subscripts in the chemical formulas, and using incorrect coefficients. It is important to double-check the balance of atoms and coefficients to avoid these errors.

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