Solve Stoichiometry Problems with Chemistry: Al & Cl2

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Discussion Overview

The discussion revolves around stoichiometry problems involving the reaction between aluminum (Al) and chlorine (Cl2) to produce aluminum chloride (AlCl3). Participants explore the correct stoichiometric ratios, the charges of the elements involved, and the interpretation of the initial quantities provided.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant initially proposes a reaction equation and calculates the production of AlCl2 based on a 5 mol Cl input, expressing uncertainty about the correctness of their approach.
  • Another participant corrects the reaction equation to 3Cl2 + 2Al → 2AlCl3 and calculates the production of AlCl3 as 3.33333 mol based on this equation, while also expressing uncertainty about potential mistakes.
  • Questions arise regarding the charge of aluminum, with one participant explaining that Al typically has a 3+ charge due to its electron configuration, while noting that a 4+ charge is theoretically possible but not favorable.
  • Several participants clarify the use of subscripts in chemical formulas and discuss formatting options for displaying chemical equations, including the use of LaTeX.
  • There is confusion about whether the initial quantity of 5 mol refers to Cl or Cl2, with participants debating the implications of this distinction on the stoichiometric calculations.
  • One participant suggests that the question lacks sufficient information to determine the correct interpretation of Cl versus Cl2, indicating that assumptions are being made about the state of chlorine.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the initial quantity of chlorine and its implications for stoichiometric calculations. There is no consensus on whether the 5 mol refers to Cl or Cl2, leading to multiple competing interpretations of the problem.

Contextual Notes

Participants highlight the ambiguity in the problem statement regarding the form of chlorine (Cl vs. Cl2), which affects the stoichiometric ratios and calculations. Additionally, there are unresolved questions about the charge of aluminum in different contexts.

PrudensOptimus
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Chemistry! Diee!

A few stoicheometry problem,.. I really feel like a noob in chem:(

A chemist has 5 mol of Cl that combines w/ Al, how many moles of solid Al Chloride is produced?

First, I set up balance eq.

5 mol Cl2 + Al ---> AlCl2

Then by ratio, 5:1, so i got 1 mole of AlCl2 is produced. Pretty sure I made a mistake. Someone help pls. (BTW, how do you determine which charge to use for Al?)
 
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The equation for the reaction between Cl2 and Al is as follows,

3Cl2 + 2Al ---> 2AlCl3 (Ionic, Al has a 3+ charge, and Cl has a 1- charge)

So 3 moles of Cl2 will react to produce 2 moles of AlCl3. Therefore if 5 mol of Cl2 is used to begin with, 3.33333 (10/3) mol of AlCl3 will be produced.


I may have made a stupid mistake somewhere. If so don't blame me, I have a hangover :smile:

Edit: How did you make the subscript numbers?
 
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When does Al have 4 charge?

subscripts are [ sub ] [ /sub ]

with no spaces.
 
Al has a 3+ charge because in order to get into the nearest noble gas configuration (1s2, 2s2, 2p6), it needs to lose 3 electrons. Theoretically it could indeed have a 4+ charge, but it would not be energetically favourable, therefore it is not likely to occur under normal conditions.
 
Originally posted by sven222

Edit: How did you make the subscript numbers?

You can also now Use Latex. Click on the image to see the code in a pop-up window.

3Cl2 + 2Al ---> 2AlCl3 (Ionic, Al has a 3+ charge, and Cl has a 1- charge)
3Cl_2+2Al \longrightarrow 2AlCl_3


Using the sub /sub flags in normal text can achieve much the same, but this looks cooler :wink:
 
Originally posted by Another God
3Cl_2+2Al \longrightarrow 2AlCl_3

You said you have 5 mols of Cl. You actually do mean Cl and not Cl2 right?
The ratio of Cl to AlCl3 is 6:2 or 3:1. So the ratio of AlCl3 to Cl is 1:3

1AlCl3/3Cl * 5Cl = 1.66666 mols AlCl3
 
Originally posted by ShawnD
You said you have 5 mols of Cl. You actually do mean Cl and not Cl2 right?
The ratio of Cl to AlCl3 is 6:2 or 3:1. So the ratio of AlCl3 to Cl is 1:3

1AlCl3/3Cl * 5Cl = 1.66666 mols AlCl3

The question didnt really give enough info. I was assuming it meant Cl gas, in which case it is Cl2. It may have meant Cl in the ionic form however, in which case you would be correct. From the info he gave though, it sounds as if it is meant to be in the gas, or Cl2, form. And also, he has Cl2 in his original post (in the equation).

Another God: Thanks, very nifty tool. Working out the commands now :)
 
Last edited:
Originally posted by Another God
You can also now Use Latex. Click on the image to see the code in a pop-up window.

3Cl2 + 2Al ---> 2AlCl3 (Ionic, Al has a 3+ charge, and Cl has a 1- charge)
3Cl_2+2Al \longrightarrow 2AlCl_3


Using the sub /sub flags in normal text can achieve much the same, but this looks cooler :wink:
It will look even better if you use itex instead of tex tags:

3Cl_2+2Al \longrightarrow 2AlCl_3

3Cl_2+2Al \longrightarrow 2AlCl_3

It decreases the font and thus allows it to put inline.
 
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