My Math Forum  

Go Back   My Math Forum > Science Forums > Chemistry

Chemistry Chemistry Forum


Thanks Tree1Thanks
  • 1 Post By studiot
Reply
 
LinkBack Thread Tools Display Modes
May 9th, 2018, 01:17 AM   #1
Member
 
Joined: May 2015
From: Australia

Posts: 77
Thanks: 7

Stoichiometry

Find the mass of iron (III) chloride that can be formed by burning 14g of iron in excess chlorine.
2Fe + 3Cl2 --> 2FeCl3

My answer:

m(Fe)=14g
Mr(Fe)=55.84g/mol
n(Fe)=14g/(55.84g/mol) =0.251mol

Ratio Fe:FeCl3
2:2
Thus, n(FeCl3) = 0.251mol

m(FeCl3) = 0.251mol x 162.147g/mol = 40.7g

My concern is that I might have calculated the Molecular Mass (Mr) incorrectly.

Would the molecular mass of 2Fe be 55.84g/mol or 111.68g/mol
Similarly, would the molecular mass of 2FeCl3 be 162.147g/mol or 324.38g/mol.
pianist is offline  
 
May 9th, 2018, 02:27 AM   #2
Senior Member
 
Joined: Jun 2015
From: England

Posts: 830
Thanks: 244

It's the atomic mass for the iron and the molecular mass for the chlorine.

But each is multiplied by the coefficient counting the number of atoms or molecules

So the coefficients are 2 for the iron and 3 for the chlorine.

Remembering that the molecular mass of chlorine = 2 x atomic mass of chlorine.
studiot is offline  
May 9th, 2018, 02:36 AM   #3
Member
 
Joined: May 2015
From: Australia

Posts: 77
Thanks: 7

So would this be the answer (taking into account the coefficient when calculating molecular mass)?

Find the mass of iron (III) chloride that can be formed by burning 14g of iron in excess chlorine.
2Fe + 3Cl2 --> 2FeCl3

My answer:

m(Fe)=14g
Mr(Fe)=55.84g/mol * 2 = 111.68 g/mol
n(Fe)=14g/(111.68g/mol) =0.125mol

Ratio Fe:FeCl3
2:2
Thus, n(FeCl3) = 0.125mol

m(FeCl3) = 0.125mol x 324.38g/mol = 40.5g
pianist is offline  
May 9th, 2018, 04:29 AM   #4
Senior Member
 
Joined: Jun 2015
From: England

Posts: 830
Thanks: 244

I think you are making this very complicated.

For every atom of iron there are 3 atoms of chlorine in the ferric chloride.

The atomic mass of iron is 55.8 so and of chlorine is 35.5.

So 55.5 grams of iron produces {55.5 + 3(35.5)} = 162 grams of ferric chloride.

So 1 gram of iron produces $\displaystyle \frac{162}{{55.5}}$ grams of ferric chloride

So 14 grams of iron produces $\displaystyle 14*\frac{{162}}{{55.5}} = 40.86$ g of ferrric chloride
studiot is offline  
May 9th, 2018, 09:54 AM   #5
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 1,810
Thanks: 724

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by pianist View Post
Find the mass of iron (III) chloride that can be formed by burning 14g of iron in excess chlorine.
2Fe + 3Cl2 --> 2FeCl3
This is the way I learned to do it. It's similar to a unit conversion.

$\displaystyle \frac{14~g~Fe}{1} \cdot \frac{1~mol~Fe}{55.845~g~Fe} \cdot \frac{2~mol~FeCl_3}{2~mol~Fe} \cdot \frac{162.195~g~FeCl_3}{1~mol~FeCl_3} = 40.66~g~FeCl_3$

The second to last factor comes from comparing the number of mols of each substance in the chemical reaction. Note how the different units cancel.

-Dan
topsquark is offline  
May 9th, 2018, 10:50 AM   #6
Senior Member
 
Joined: Jun 2015
From: England

Posts: 830
Thanks: 244

Thank you Dan for correcting my arithmetic.


My point is that you don't need the reaction equation at all, just the formula for ferric chloride.
The question says in "excess chlorine" so all the iron is used up.
Thanks from topsquark
studiot is offline  
May 9th, 2018, 04:33 PM   #7
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 1,810
Thanks: 724

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by studiot View Post
Thank you Dan for correcting my arithmetic.


My point is that you don't need the reaction equation at all, just the formula for ferric chloride.
The question says in "excess chlorine" so all the iron is used up.
True. I'm known for doing things the hard way.

-Dan
topsquark is offline  
Reply

  My Math Forum > Science Forums > Chemistry

Tags
stoichiometry



Thread Tools
Display Modes






Copyright © 2018 My Math Forum. All rights reserved.