What mass of P4 is required to combine with 0.398 g of Cl2?

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Discussion Overview

The discussion revolves around a stoichiometry problem involving the reaction of phosphorus (P4) with chlorine (Cl2) to produce phosphorus trichloride (PCl3). Participants are attempting to calculate the mass of P4 required to react with a given mass of Cl2 and the resulting mass of PCl3 produced. The scope includes homework-related problem-solving and mathematical reasoning.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses difficulty in applying stoichiometric factors to solve the problem, indicating confusion with the process.
  • Another participant corrects the atomic mass values for chlorine and phosphorus, emphasizing the need for accurate data.
  • Participants discuss the calculation of moles from the given mass of Cl2 and the conversion to grams for P4, with varying degrees of clarity and correctness.
  • There is a calculation presented for the molecular weight of P4, leading to a derived mass based on moles calculated earlier.
  • Participants explore the relationship between moles of P4 and moles of PCl3 produced, with some confusion about the stoichiometric coefficients from the balanced equation.
  • One participant suggests using proportions as a foundational concept for solving stoichiometry problems, indicating a pedagogical approach to the discussion.
  • Final calculations for the mass of PCl3 are discussed, with one participant arriving at a specific mass based on previously calculated moles.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to the stoichiometry problem initially, with some confusion and corrections occurring throughout the discussion. However, there is a general agreement on the final calculations for the mass of P4 and PCl3 based on the stoichiometric relationships.

Contextual Notes

Some calculations and assumptions are not fully resolved, including the correct application of stoichiometric factors and the transition between moles and grams. The discussion reflects varying levels of understanding of stoichiometry among participants.

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Homework Statement


I have tried and tried and I can't figure out the factors to use on this:
a) What mass of P4 is required to combine with 0.398 grams (g) of Cl2 to produce PCl3?
b) What mass of PCl3 results when 0.398 g of Cl2 is combined with the calculated mass of P4?
I know that the equation is already balanced but I am not sure how to write the factors to get the answers. I am teaching myself, I'm not in chemistry class, so please be patient, thank you. I still haven't gotten the hang of using factors for stoichiometry problems.

Homework Equations


Equation to use: P4(s) + 6Cl2(g) ---> 4PCl3 (l) Unknown: P4 Known: 0.398g Cl2
I know that the atomic mass of Cl2 is 35.453 and the atomic mass of P4 is 30.973

The Attempt at a Solution


I tried using these factors but none of them yielded the right answers: 1 mole P4/6 mole Cl2, 1 mole PCl3/137.3 g PCl3, and 1 mole Cl2/70.906 g Cl2 but none of them work. Like I said, I tried to solve it using the factors but I cannot figure out which to use, they are confusing to me.
I started with 0.00561 moles of Cl2 times 1 mole P4/6 moles Cl2 and I crossed out the Cl2 which would give 0.000935 P4 which is not the answer. The answers are supposed to be 0.116 g P4 and 0.514 g PCl3
 
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science_rules said:
I know that the atomic mass of Cl2 is 35.453 and the atomic mass of P4 is 30.973
That's not correct. The atomic mass of Cl is 35.453 u and the atomic mass of P is 30.973 u.
 
I just forgot to take the subscripts off
 
science_rules said:
I started with 0.00561 moles of Cl2 times 1 mole P4/6 moles Cl2 and I crossed out the Cl2 which would give 0.000935 P4 which is not the answer.

0.000935 of what? Meters? Hours?
 
0.000935 moles
 
science_rules said:
0.000935 moles
Right. So what's the molecular weight of P4? So how many grams does 0.000935 moles of P4 represent?
 
30.97 x 4 = 123.88 amu molecular weight, 0.028 grams in 0.000935 moles of P4 because 0.000935 moles times 30.97 =
0.028 grams
 
science_rules said:
30.97 x 4 = 123.88 amu molecular weight, 0.028 grams in 0.000935 moles of P4 because 0.000935 moles times 30.97 =
0.028 grams
##0.000935\, moles \times 123.88 \frac{grams}{mole}=??##
 
= 0.116 grams P4
 
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  • #10
Now: b) What mass of PCl3 results when 0.398 g of Cl2 is combined with the calculated mass of P4? Do I just multiply 0.398 g of Cl2 times 0.116 grams of P4?
 
  • #11
oops, maybe not
 
  • #12
science_rules said:
Now: b) What mass of PCl3 results when 0.398 g of Cl2 is combined with the calculated mass of P4? Do I just multiply 0.398 g of Cl2 times 0.116 grams of P4?
That will give you something with units of grams2. From your balanced reaction equation, how many moles of PCl3 are produced for every mole of P4 reacted?
 
  • #13
1 mole?
 
  • #14
I mean 4 moles
 
  • #15
science_rules said:
I mean 4 moles
So, if 0.000935 moles of P4 react, how many moles of PCl3 are produced?
 
  • #16
4 moles times 0.000935 moles = 0.00374?
 
  • #17
0.00374 moles of PCl3
 
  • #18
science_rules said:
0.00374 moles of PCl3
Correct. You're on a roll. What is the mass of 0.00374 moles of PCl3?
 
  • #19
2.723 grams?
 
  • #20
science_rules said:
2.723 grams?
Where did you get that from? What is the molecular weight of PCl3 in grams per mole?
 
  • #21
molecular weight of PCl3 = 30.97376 + 35.453X3 = 30.97376 + 106.359 = 137.33276 grams PCl3/mole
 
  • #22
science_rules said:
molecular weight of PCl3 = 30.97376 + 35.453X3 = 30.97376 + 106.359 = 137.33276 grams PCl3/mole
So...?
 
  • #23
Do I divide 0.398 g Cl2 by the 70.9064 g Cl2? which would equal 0.00561 moles?
 
  • #24
Please pay attention to what you are doing - the step you are doing now has nothing to do with chlorine, you have already correctly calculated how many moles of PCl3 are produced, and you were trying to find the mass of

science_rules said:
0.00374 moles of PCl3

Why do you step back to chlorine?
 
Last edited:
  • #25
My advice:
1 draw and use pictures as I mentioned in a recent post
2
epenguin said:
This is a question in very starting chemistry. If this was a problem, if you needed a formula, or needed the indications of a teacher, if you were stuck by that question, and you are starting chemistry, let me suggest you do yourself a big favour:

Revise SIMPLE PROPORTIONS!

Whether it needs ten minutes and then it comes back, or whether you take a school arithmetic book and work through numbers of examples, do it till it is easy and obvious.

Because ALL the quantitative problems you will meet and be asked to solve for the first years in Chenistry will be really about nothing but proportions,
 
  • #26
0.00374 moles of PCl3 times 137.33 grams per mole of PCl3 gives you how many grams of PCl3?
 
  • #27
0.514 g of PCl3
 
  • #28
epenguin said:
My advice:
1 draw and use pictures as I mentioned in a recent post
2
thank you
 
  • #29
Chestermiller said:
0.00374 moles of PCl3 times 137.33 grams per mole of PCl3 gives you how many grams of PCl3?
Thank you for your help!
 

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