Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Kdub on October 10, 2005, 10:52:44 PM
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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?
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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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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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1. I calculated that moles Al = moles H2 x 2/3 = 0.0970 mol
Can you explain where have you got 2/3 from?
Concentration of original stock solution = 0.987mol / 0.847L = 1.17 M
No idea what you trying to do here.
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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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
No idea what you trying to do here.
trying to find original concentration
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
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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.
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 (http://www.chembuddy.com/?left=concentration&right=percentage-to-molarity).
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 (http://www.chembuddy.com/?left=concentration&right=dilution-mixing) lecture).
You may also try CASC - although you will only get the result, not the way it was calculated.
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without using the calculator I got 0.1332 L required but i still have a feeling that's wrong.
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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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Ok thanks, got #2 out of the way!
now to get #1..
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*Ignore me, I am impatient* for first question!
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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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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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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?
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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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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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need to get this question!
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So start from both correct equations, you know how to do the math - DO IT!
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I keep getting moles in the negatives...but I don't know where my mistake is:
55.85x + 26.98y = 3.753
55.85(0.1456 - 2/3y) + 26.98y = 3.753
8.132 - 37.233y + 26.98y = 3.753
-10.253y = -4.379
y = 0.4271
x + 2/3y = 0.1456
x + 2/3(0.4271) = 0.1456
x + 0.2847 = 0.1456
x = -0.1391 ... theres the mistake... ???
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Second equation should be
x + 3/2y = 0.1456
Sorry, you are making mistakes faster than I can correct them.
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44.5% finally ;D