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Topic: Stock solution and percentage by mass  (Read 16735 times)

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Kdub

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Stock solution and percentage by mass
« on: October 10, 2005, 10:52:44 PM »
Just need a few leading tips to get started on these 2 questions:

1. A 3.753g sample of a mixture of Fe and Al is treated with excess HCl (aq). If 0.1456 moles of H2 are obtained, then what is the percentage by mass of Fe in the original mixutre?
Reactions:
Fe(s) 2HCl(aq) --> FeCl2(aq) + H2(g)
2Al (s) + 6HCl (aq) --> 2AlCl3(aq) + 3H2(g)

Molar Masses: Al=26.95g/mol                  Fe=55.85g/mol

So far I have that xFe + yAl = 3.753g. I just can't figure out how to get moles of Fe and moles of Al from that.


2. An aqueous stock solution is 36.0% HCl by mass and its density is 1.18g/mL. What volume of this solution is required to make 1.25L of 1.25mol/L HCl(aq)?
Molar Masses: HCl=36.458g/mol             H2O=18.016g/mol

What I have is:
Assume a 100g sample...
Moles HCl = 36g / 36.458
               = 0.987mol

Moles H2O = 64g / 18.016
                = 3.55mol

Volume of Solution = 100g / 1.18g/mL / 1000
                           = 0.0847 L

From there I don't know if to calculate concentration of initial volume is to just do moles HCl / volume, or if i need to calculate mole fraction and use that number to divide by volume?
« Last Edit: October 11, 2005, 12:16:09 AM by Kdub »
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Offline AWK

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Re:Stock solution and percentage by mass
« Reply #1 on: October 11, 2005, 02:33:22 AM »
1. Hint
moles_Al = mass_Al / 26.95
note! - 1 mole Al form 1.5 moles of H2

2. Hint
 1.25 L of 1.25 M solution contains 1.25x1.25 moles of HCl
You calculated correctly moles of HCl and volume for 100 g of stock solution
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Kdub

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Re:Stock solution and percentage by mass
« Reply #2 on: October 11, 2005, 01:16:53 PM »
Thanks for the hints, so ...
1. I calculated that moles Al = moles H2 x 2/3
                                        = 0.0970 mol
Then mass Al = moles x molar mass
                    = 2.616g
Therefore mass Fe in sample = 3.753g - 2.616g
                                          = 1.137g

So percent by mass of Fe = (1.137g / 3.753g) x 100%
                                    = 30.3%


2. moles of HCl in concentrated solution = 1.25 x 1.25
                                                         = 1.563mol

Concentration of original stock solution = 0.987mol / 0.847L
                                                        = 1.17 M

I ended up with 1.34L .Does that seem right?
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Offline Borek

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Re:Stock solution and percentage by mass
« Reply #3 on: October 11, 2005, 01:27:00 PM »
Quote from: Kdub on October 11, 2005, 01:16:53 PM
1. I calculated that moles Al = moles H2 x 2/3 = 0.0970 mol

Can you explain where have you got 2/3 from?

Quote
Concentration of original stock solution = 0.987mol / 0.847L = 1.17 M

No idea what you trying to do here.

Quote
I ended up with 1.34L .Does that seem right?

No. Although digits are OK, just the decimal point is in the wrong place.
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Kdub

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Re:Stock solution and percentage by mass
« Reply #4 on: October 11, 2005, 01:36:18 PM »
Quote
Can you explain where have you got 2/3 from?
for every 2 moles of Al there are 3 moles of H2 .. its molar ratio

Quote
No idea what you trying to do here.
trying to find original concentration

Quote
No. Although digits are OK, just the decimal point is in the wrong place.
can you point out where i should start correction instead of just saying its wrong. thx
« Last Edit: October 11, 2005, 01:37:49 PM by Kdub »
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Offline Borek

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Re:Stock solution and percentage by mass
« Reply #5 on: October 11, 2005, 02:13:31 PM »
Quote from: Kdub on October 11, 2005, 01:36:18 PM
for every 2 moles of Al there are 3 moles of H2 .. its molar ratio

So you are assuming that all hydrogen is from Al reaction? What about Fe?

Initially you assumed masses of Al and Fe to be x and y. Try to convert these masses to numbers of moles and express hydrogen moles through these values.

Quote
trying to find original concentration

What original concentration? There are two concentrations GIVEN in the question.

Hmmm... perhaps you meant that you are trying to convert units. Try this percentage to molarity lecture.

Quote
can you point out where i should start correction instead of just saying its wrong.

That's not easy if I don't understand what you did - and how.

Convert % concentration to molarity, then use CV=const (read this dilution and mixing lecture).

You may also try CASC - although you will only get the result, not the way it was calculated.
« Last Edit: October 11, 2005, 02:15:52 PM by Borek »
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Kdub

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Re:Stock solution and percentage by mass
« Reply #6 on: October 11, 2005, 03:00:44 PM »
without using the calculator I got 0.1332 L required but i still have a feeling that's wrong.
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Offline Borek

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Re:Stock solution and percentage by mass
« Reply #7 on: October 11, 2005, 03:25:07 PM »
Quote from: Kdub on October 11, 2005, 03:00:44 PM
without using the calculator I got 0.1332 L required but i still have a feeling that's wrong.

It is pretty close to the real value shown by CASC (134 mL) so it can be OK (CASC takes also final density into account). Show your calculations step by step if you are still in doubt.
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Kdub

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Re:Stock solution and percentage by mass
« Reply #8 on: October 11, 2005, 04:03:33 PM »
Ok thanks, got #2 out of the way!

now to get #1..
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Kdub

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Re:Stock solution and percentage by mass
« Reply #9 on: October 11, 2005, 11:42:48 PM »
*Ignore me, I am impatient* for first question!
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Kdub

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Re:Stock solution and percentage by mass
« Reply #10 on: October 12, 2005, 12:06:18 PM »
ok i think i finally got number one, just need some one to tell me its okay.

Let x be #of moles of Fe in original sample.
Let y be #of moles of Al in original sample.

x mol Fe --> x mol H2
y mol Al --> 3y mol H2            
Therefore x + 3y = 0.1456 (equation 1)

Also,
x moles Fe = 55.85x x grams Fe
y moles Al = 26.98y  x grams Al
Therefore 55.85x + 26.98y = 3.753 (equation 2)

Re-arrange equation 1:
x = 0.1456 - 3y

Then sub into 2:
55.85(0.1456 - 3y) + 26.98y = 3.753
8.132 - 167.55y + 26.98y = 3.753
-140.57y = -4.379
           y = 0.03155

Therefore: x + 3(0.03115) = 0.1456
                                    x = 0.05215 mol

So then, mass Fe in mixture = 0.05215 mol x 55.85 g/mol
                                         = 2.913g

And percent by mass = (2.913 / 3.753) x 100%
                              = 77.7%

PHEW!
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Offline Borek

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Re:Stock solution and percentage by mass
« Reply #11 on: October 12, 2005, 01:05:00 PM »
Quote from: Kdub on October 12, 2005, 12:06:18 PM
x mol Fe --> x mol H2
y mol Al --> 3y mol H2

Not exactly. Check balanced reaction equations.

Everything else looks OK.
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Kdub

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Re:Stock solution and percentage by mass
« Reply #12 on: October 12, 2005, 01:20:34 PM »
oops!! y moles Al = 2/3 moles H2

will that affect only equation 1, or both? like, can i just use the 0.03115 i got and multiply it by 2/3 instead of 3?
« Last Edit: October 12, 2005, 01:21:43 PM by Kdub »
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Offline Borek

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Re:Stock solution and percentage by mass
« Reply #13 on: October 12, 2005, 01:29:52 PM »
Quote from: Kdub on October 12, 2005, 01:20:34 PM
will that affect only equation 1, or both? like, can i just use the 0.03115 i got and multiply it by 2/3 instead of 3?

0.03115 was calculated using both equations - first included - so it is wrong. Second equation is OK.
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Kdub

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Re:Stock solution and percentage by mass
« Reply #14 on: October 12, 2005, 01:45:07 PM »
so for the second equation would it be 26.98(2/3y) and then leave it to the end where we have 2/3y = 0.03115 to get y=0.02077 or is it just 17.987y then work your way through
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