How to Calculate Mass in Stoichiometry Problems?

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Discussion Overview

The discussion revolves around calculating mass in stoichiometry problems, specifically involving aluminum chloride and sodium hydroxide in a double displacement reaction, as well as determining the mass of copper(II) sulfate pentahydrate needed for a specific solution concentration. The scope includes theoretical calculations and practical applications in chemistry.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a stoichiometry problem involving aluminum chloride and sodium hydroxide, asking for the mass of aluminum hydroxide produced and the excess reagent remaining.
  • Another participant suggests converting volume to liters and multiplying by concentration to find moles of copper(II) sulfate pentahydrate, then calculating grams using molecular weight.
  • Several participants discuss the molecular weight of copper(II) sulfate pentahydrate, with differing calculations leading to values around 249.71 g/mol and 250 g/mol.
  • One participant provides a calculation for the moles of aluminum chloride and sodium hydroxide, questioning the method for determining the remaining mass of sodium hydroxide after the reaction.
  • Another participant outlines the chemical reaction and provides a method for calculating the mass of sodium hydroxide that reacts with aluminum chloride.
  • Some participants express uncertainty about the calculations and suggest that understanding how to balance equations is crucial for solving stoichiometry problems.
  • One participant confirms their calculation of the remaining mass of sodium hydroxide as approximately 1.5 g.

Areas of Agreement / Disagreement

Participants express differing views on the correct molecular weight of copper(II) sulfate pentahydrate and the calculations related to the stoichiometry problems. There is no consensus on the accuracy of the molecular weight or the methods used to solve the problems.

Contextual Notes

Some calculations depend on the accuracy of molecular weights and the proper balancing of chemical equations, which remain points of contention among participants.

Samantha
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1. Aluminum chloride (15.0g) + sodium hydroxide (15g) = double displacement reaction.Predict mass of aluminum hydroxide produced. What mass of the excess reagent remains unreacted?

2. What mass of copper(II) sulfate pentahydrate is needed to prepare 150.0 mL of a 0.125mol/L solution?

How do I find them?
 
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2. convert 150 ml to L then multiply Volume * Concentration. Therefore = (0.150 L)(0.125 mol/L). This will get you the number of moles of copper (II) sulfate pentahydrate. Then multiply the Molecular Weight of copper (II) sulfate pentahydrate by the number of moles and you will get the number of grams. Hope that helps.
 
does the Molecular Weight of copper (II) sulfate pentahydrate = 164.21g/mol?
 
Not sure, if you calculated it using the molecular weights then it's probably right.
 
MusicMonkey said:
Not sure, if you calculated it using the molecular weights then it's probably right.

copper(II) sulfate pentahydrate

This is what I did:

(molar mass of Cu * 2 ) + ( molar mass of S) + (molar mass of H * 5
 
Cu+S+O*4+5*(H*2+O)

The molecular formula is CuSO4(H2O)5
 
Therefore your molecular weight is incorrect. Try again.
 
MusicMonkey said:
Therefore your molecular weight is incorrect. Try again.


I re-calculated it and it came out to 249.71g/mol??
 
Samantha said:
I re-calculated it and it came out to 249.71g/mol??

It should come out 250.

What about the first problem??Can u do it...?

Daniel.
 
Last edited:
  • #10
63.546+32.066+15.9994*4+5*(1.00794*2+15.9994)
I believe it should come out to be 249.686
 
  • #11
1. Is this right:

Mol of AlCl3 = 0.112mol
Mol of NaOH = 0.125mol

mass = (0.112mol)(78.03g/mol)
= 8.74g

b)do I multiply 0.125 mol by the mols of 3NaCl. Then subtract 8.74 from the anwser?
 
  • #12
Depending on the number of the significant figures used I believe that your answer should be accurate enough.
 
  • #13
The final answer for question 2 should be 4.68 g.
 
  • #14
MusicMonkey said:
63.546+32.066+15.9994*4+5*(1.00794*2+15.9994)
I believe it should come out to be 249.686

No,no,no,no,no.No such precision required whatsoever.YOUR FINAL ANSWER IS APPROXIMABLE TO 250... :-p

Daniel.
 
  • #15
Samantha said:
1. Is this right:

Mol of AlCl3 = 0.112mol
Mol of NaOH = 0.125mol

mass = (0.112mol)(78.03g/mol)
= 8.74g

b)do I multiply 0.125 mol by the mols of 3NaCl. Then subtract 8.74 from the anwser?

The remaining mass is of NaOH indeed & is roughly 1.5g

Daniel.
 
  • #16
dextercioby said:
The remaining mass is of NaOH indeed & is roughly 1.5g

Daniel.

How did you get that?
 
  • #17
1 mole AlCl_{3}----------------->133.5g
1 mole NaOH-------------------------> 40g

The chemical reaction:
[tex]3NaOH+AlCl_{3}\rightarrow Al(OH)_{3}+3NaCl[/tex]

So
3 moles NaOH/120g react with 1mole AlCl_{3}/133.5 g.
x g NaOH react with 15g of AlCl_{3}

Find "x" & subtract it from 15.

Daniel.
 
  • #18
dextercioby said:
1 mole AlCl_{3}----------------->133.5g
1 mole NaOH-------------------------> 40g

The chemical reaction:
[tex]3NaOH+AlCl_{3}\rightarrow Al(OH)_{3}+3NaCl[/tex]

So
3 moles NaOH/120g react with 1mole AlCl_{3}/133.5 g.
x g NaOH react with 15g of AlCl_{3}

Find "x" & subtract it from 15.

Daniel.

:confused:
 
  • #19
Samantha, before solving problems on stoichiometry, you need to learn how to write and balance equations. I suggest you first work on this, and get your basics clear. It may take a while but it's very important, and essential to your being able to solve any kind of chemistry problem.
 
  • #20
Gokul43201 said:
Samantha, before solving problems on stoichiometry, you need to learn how to write and balance equations. I suggest you first work on this, and get your basics clear. It may take a while but it's very important, and essential to your being able to solve any kind of chemistry problem.
I got the same equation but I don't get the calculations..I should work on that.
 
  • #21
I tried it and I got 1.5g :smile:

Thank you
 

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