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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: purple_rose on March 16, 2008, 05:14:11 pm

Title: Heinemann questions i had trouble with
Post by: purple_rose on March 16, 2008, 05:14:11 pm: 1. What mass of barium chloride(CaCl2) will remain after 15g sample of the hydrated salt BaCl2.2H20 is heated to drive off all of the water? (The answer is 12.8g, i keep getting 12.4g)

2. A .5g sample of sodium sulfate (Na2SO4) and .5g sample of aluminium sulfate (Al2(SO4)3 ) were dissolved in a volume of water and excess barium chloride added to preciptate barium sulfate. What was the totoal mass of barium sulfate produced?

(i keep getting 1.02g for this one but the answer is 1.85g)

3. 1. The chlorine in a .63g sample of chlorinate pesticide, DDT (C14H9Cl5), is precipitated as silver cloride. What mass of silver chloride is formed?

4. A precipitate of Fe2O3, of mass 1.43g, was obtained by treating a 1.5L sample of bore water. What was the concentration of iron, in molL-1, in the water? (the answer is 1.012molL-1, but i keep getting 1.013molL-1, i noe it seems pedantic but i don't noe y it's higher)
Title: Re: Heinemann questions i had trouble with
Post by: Collin Li on March 16, 2008, 05:32:36 pm: Question 1

$\mbox{m(BaCl}_2\mbox{)} = 15.0\mbox{ g} \times \frac{137.3 + 2(35.5)\mbox{ gmol}^{-1}}{137.3 + 2(35.5) + 2(18.0)\mbox{ gmol}^{-1}} = 12.8\mbox{ g}$

(The $18.0$ comes from the molar mass of water)

How are you doing it?
Title: Re: Heinemann questions i had trouble with
Post by: purple_rose on March 16, 2008, 05:40:21 pm: oic, i must of messed up the molar masses, thanks
Title: Re: Heinemann questions i had trouble with
Post by: Collin Li on March 16, 2008, 05:44:16 pm: Question 2

Find out the amount of sulfate ions present:

$\mbox{n(SO}_4\mbox{}^{2-}\mbox{)} = \frac{0.5}{2(23.0) + 96.1} + 3\times\frac{0.5}{2(27.0) + 3(96.1)} = 0.00790\mbox{ mol}$

(The $96.1$ comes from the molar mass of a sulfate ion)

$\implies \mbox{m(BaSO}_4\mbox{)} = 0.00790 \times (137.3 + 96.1) = 1.84\mbox{ g}$

I kept all the digits in my calculator. The next decimal places were: "1.84403..." so it's easy to see how the book was off by 0.01. They probably rounded off their calculations too early, or perhaps they used more accurate molar masses - I always use 1 decimal place, because that is how VCAA does it. A difference not worth worrying about.

Similarly, I would not bother with the difference you got in Question 4. The people who write the solutions in the books tend to round off their calculations in the middle, which reduces the accuracy of their final answer.
Title: Re: Heinemann questions i had trouble with
Post by: Collin Li on March 16, 2008, 05:52:25 pm: Question 3

These questions should be easy (similar to Question 1). You should be using the fact that elements are not created or destroyed (unless it is a nuclear reaction).

This means you can say that the amount of chlorine atoms in DDT is equal to the amount of chlorine atoms in silver chloride.

$\mbox{m(AgCl)} = 5\times \frac{\mbox{m(DDT)}}{\mbox{M(DDT)}} \times \mbox{M(AgCl)}$

I'll let you have a go at the numbers. By uppercase "M", I mean the molar masses.
Title: Re: Heinemann questions i had trouble with
Post by: purple_rose on March 16, 2008, 05:53:50 pm: um....for question 2 why is there a +3?
Title: Re: Heinemann questions i had trouble with
Post by: Collin Li on March 16, 2008, 05:57:48 pm: It's 3 multiplied by the last fraction, because that last fraction represents the amount of aluminium sulfate, but there are 3 sulfate ions per aluminium sulfate.
Title: Re: Heinemann questions i had trouble with
Post by: purple_rose on March 16, 2008, 06:06:16 pm: For question 3, is it possible to write an equation for wot is going on?

So this is how u do it, correct me if i'm wrong
1. so u find the mole of DDT which is .00356
2. you then times by 5 becos theres 5 Cl which is .017797
3. you then times by M of silver chlorine (which is 143.32) to get the mass of AlCl, i got .08898, but the back of says the answer is 1.3g

"because that last fraction represents the amount of aluminium sulfate"...... i don't understand wot u mean
Title: Re: Heinemann questions i had trouble with
Post by: Collin Li on March 16, 2008, 06:11:42 pm: Yeah, what you said in words is spot on, but I think your numbers are wrong:

Using:
$\mbox{M(DDT)} = 14(12.0) + 9(1.0) + 5(35.5) = 354.5\mbox{ gmol}^{-1}$

$\mbox{M(AgCl)} = 107.9 + 35.5 = 143.4\mbox{ gmol}^{-1}$

$\mbox{m(AgCl)} = 5\times \frac{\mbox{m(DDT)}}{\mbox{M(DDT)}} \times \mbox{M(AgCl)}$

$= 5\times \frac{0.63}{354.5} \times 143.4 = 5\times 0.00178 \times 143.4 = 1.27\mbox{ g}$
Title: Re: Heinemann questions i had trouble with
Post by: Collin Li on March 16, 2008, 06:15:32 pm: Quote from: purple_rose on March 16, 2008, 06:06:16 pm
"because that last fraction represents the amount of aluminium sulfate"...... i don't understand wot u mean

Quote
$\mbox{n(SO}_4\mbox{}^{2-}\mbox{)} = \frac{0.5}{2(23.0) + 96.1} + 3\times\frac{0.5}{2(27.0) + 3(96.1)} = 0.00790\mbox{ mol}$

This is:

$\mbox{n(SO}_4\mbox{}^{2-}\mbox{)} = \mbox{n(Na}_2\mbox{SO}_4\mbox{)} + 3\times\mbox{n(Al}_2\mbox{(SO}_4\mbox{)}_3\mbox{)}$

The fractions in the top equation with numbers are just the masses divided by the molar masses (which give you the amount of moles).
Title: Re: Heinemann questions i had trouble with
Post by: purple_rose on March 16, 2008, 06:17:57 pm: ooooooooooooooooooooooooooo. i noe wot i did wrong, i timed it by 5 twice,
i get it now, awesome

thanks coblin
Title: Re: Heinemann questions i had trouble with
Post by: Collin Li on March 16, 2008, 06:18:44 pm: No problem :)
Title: Re: Heinemann questions i had trouble with
Post by: lanvins on March 24, 2008, 05:18:49 pm: More heinemann question help- Mostly half equations questions i had trouble with

1. 2S2O4 -2(aq) + H20(l) = S2O3 -2(AQ) + 2HSO3 - (aq)

Write the ionic half equations for the oxidation and reductions that occur when sodium dithionite is mixed with water

2. In dry cells commonly used in torches, an electric current is produced from the reaction of zinc metal with MnO2. During this reaction, Zn2+ ions and Mn2O3 are formed.

Calculate the mass of zinc that would be needed to react completely with 8.g of MnO3 in a dry cell.
I got 6g but thats wrong according to the book

3. The thermite process can be used to weld lengths of railway track together. A mould placed over the ends of the rails to be joined is filled with a charge of aluminium powder and iron(iii) oxide. When the mixture is ignited, a redox reaction occurs to form molten iron, which joins the rails together.
a. Write a half equation for the conversion of iron(iii) oxide to metallic iron.
i got Fe2O3 + 6H+ +6e- = 2Fe + 3H20- but thats wrong according to the book

b. Write the overall equation for the thermite process.
Title: Re: Heinemann questions i had trouble with
Post by: Mao on March 24, 2008, 06:00:08 pm: 1) $2S_2O_4^{-2}(aq)+H_2O(l)\rightarrow S_2O_3^{-2}(aq)+2HSO_3 ^-$

reduction:
$S_2O_4^{-2}(aq)+2H^+(aq)+2e^- \rightarrow S_2O_3^{-2}(aq)+H_2O(l)$

oxidation:
$S_2O_4^{-2}(aq)+2H_2O(l)\rightarrow 2HSO_3^-(aq)+2H^+(aq) +2e^-$

2) the equation will be:

$Zn(s)+2MnO_2(s)+2H^+(aq)\rightarrow Mn_2O_3(s)+H_2O(l)+Zn^{2+}(aq)$

i'm assuming you meant 8.0g of MnO₂:

$n(Zn)=\frac{1}{2}n(MnO_2)=\frac{1}{2}\cdot \frac{8.0}{54.9+16.0 \times 2}=0.0460mol$

$m(Zn)=0.0460mol \times 65.4 g mol^{-1}=3.0g$

3) In the mould, there are no hydronium ions present. In fact, the reaction does not involve (or occur in the presence of) water. So in this case, the oxygen is balanced by $O^{-2}$ instead of $H_2O$

a) $Fe_2O_3(s)+6e^-\rightarrow 2Fe(s)+3O^{2-}(s)$

b) $Fe_2O_3(s)+2Al(s)\rightarrow Al_2O_3(s)+2Fe(s)$
Title: Re: Heinemann questions i had trouble with
Post by: lanvins on March 24, 2008, 06:06:37 pm: yea i get that....so ur allowed to use S204-2 twice?....i thought i had to use the water
Title: Re: Heinemann questions i had trouble with
Post by: Mao on March 24, 2008, 06:44:02 pm: Quote from: lanvins on March 24, 2008, 06:06:37 pm
yea i get that....so ur allowed to use S204-2 twice?....i thought i had to use the water
notice the oxidation numbers in $S_2O_4^{2-}, S_2O_3^{2-}, HSO_3^-$ , the number for sulfur is +3, +2 and +5.
since the latter two sourced from the same compound (the first one), that means some of it must have gone through oxidation, and some gone through reduction.
Title: Re: Heinemann questions i had trouble with
Post by: lanvins on March 24, 2008, 06:51:39 pm: Sorry to bother u, but i'm still a little confused about Q3 like what are hydronium ions and how did u know to balance it with O2- (where did it say in the question)? Also, how did u do Q3 b, like where did Al2O3 come from?
Title: Re: Heinemann questions i had trouble with
Post by: Mao on March 24, 2008, 07:57:55 pm: in the thermite reaction, everything happens under that mold, and the question did not say "there are also hydronium ions in the paste", so you cannot assume that there are H⁺ ions available for this reaction.
as a result, you'd balance the oxides with oxide ions, O^2-

in a reaction, say, involving the reduction of permanganate (which involves H⁺ ions), the question would say that it is an "acidified solution", meaning you can follow the KOHES rule

if you had written the other half of the equation:
$Al(s)\rightarrow Al^{3+}(s)+3e^{-}$
and combined the two half equation together, it becomes apparent that Al₂O₃ is formed from the two aluminium ions and the three oxide ions.
Title: Re: Heinemann questions i had trouble with
Post by: lanvins on March 24, 2008, 08:16:14 pm: i promise this is the last time i'm gonna bother u.
So tell me if this is right:
1. So whenever it says its a thermite reaction, you balance it with O2-
(r there a list of these reactions i should noe?)

Quote from: Mao on March 24, 2008, 07:57:55 pm
if you had written the other half of the equation:
$Al(s)\rightarrow Al^{3+}(s)+3e^{-}$
and combined the two half equation together, it becomes apparent that Al₂O₃ is formed from the two aluminium ions and the three oxide ions.

2. Two Half equations
Al(s) = Al3+(s) + 3e-
FeO3(s) + 6e- = 2Fe(s) 3O2-

Getting rid of the e-
2Al = 2Al3+ + 6e-
Fe2O3 + 6e- = 2Fe + 3O2-

Overall equation
2Al + Fe2O3 = 2Al 3+ + 3O 2-+ 2Fe
Title: Re: Heinemann questions i had trouble with
Post by: Mao on March 24, 2008, 08:25:31 pm: yes, kind of, not really

1: you can only use $H^+$ when there are some. In a question that you will be using it, the question will tell you that the solution is "acidified" so you can use it. Those are generally only applicable to Mn and V compounds.

2. The reaction is actually a lot more spontaneous than that, like a "swap" of metal ions. Half equations is useful to determine what is happening, but it is not absolute.
Title: Re: Heinemann questions i had trouble with
Post by: lanvins on March 25, 2008, 08:23:29 am: r u sure they didn't just simplify the H in Fe2O3 +6e- +6 H+ --> 2 Fe + 3 H2O to get Fe2O3(s) + 6e- --> 2Fe(l) + 3O 2- (s)?

(this question is so retarded)
Title: Re: Heinemann questions i had trouble with
Post by: Mao on March 25, 2008, 10:07:54 pm: Quote from: lanvins on March 25, 2008, 08:23:29 am
r u sure they didn't just simplify the H in Fe2O3 +6e- +6 H+ --> 2 Fe + 3 H2O to get Fe2O3(s) + 6e- --> 2Fe(l) + 3O 2- (s)?

(this question is so retarded)
unless you can explain how some hydronium ions miraculously appeared inside the mold, then yes :D
Title: Re: Heinemann questions i had trouble with
Post by: lanvins on March 25, 2008, 11:17:08 pm: what do you think the chances are of a question of that kind being in the exam (cause i would be f...ked)?

It would be nice if the textbok mentioned something about these sepcial exceptions for balancing half equations.
Title: Re: Heinemann questions i had trouble with
Post by: Mao on March 26, 2008, 10:20:10 am: I have not seen thermite reactions in exams, but just to be safe, read the questions carefully and determine whether there will be H⁺ available
The key is to not "create" something the question has not given you. That can be a deciding factor between a high achiever and a very high achiever.

and maybe if you are unsure about that, memorise the thermite reaction formula.