VCE Chemistry Question Thread

[nhmn0301]

I asked my teacher about this and he said that the 3O² isn't supposed to be there and now I'm pretty confused. I was wondering if someone could show me step by step to find this half equation.

Btw, if this is a reduction half equation, why aren't the electrons on the left side?

Thanks :-)

Here is my answer, step by step, ( I'm too lazy to include states in each step, but I will at the end of my answer)
Step 1: you balance the KEY elements
Fe2O3 -> 2Fe
Step 2: balancing the O by adding water ( as you can see, there are 3O on the LHS, hence, you must add 3H2O to the RHS)
Fe2O3 -> 2Fe + 3H2O
Step 3: balance the H+ , on the  RHS, you see that there are 6H+, hence, to balance this, add 6H+ on the LHS
Fe2O3 + 6H+ -> 2Fe + 3H2O
Step 4: balancing all the charges by adding electrons. As you can see, on the LHS:
Fe2O3 + 6H+ -> 2Fe + 3H2O
(   0     +   6   ->  0    +  0     )      ( This step is actually finding the Oxidation number of each compound, add them together and find out where you can put the electrons to balance both side of the equation)
Hence, I see that there are 6+ on the LHS and 0 on the RHS, I would add 6e- to the LHS to balance the charges since 6 + (-6) = 0 ( = RHS)

Fe2O3 + 6H+ 6e- -> 2Fe + 3H2O
Step 5: include stages
Fe2O3(s) + 6H+(aq) + 6e- -> 2Fe(l) + 3H2O(l)
Hope this helps, let me know if there is any errors!

[lzxnl]

Here is my answer, step by step, ( I'm too lazy to include states in each step, but I will at the end of my answer)
Step 1: you balance the KEY elements
Fe2O3 -> 2Fe
Step 2: balancing the O by adding water ( as you can see, there are 3O on the LHS, hence, you must add 3H2O to the RHS)
Fe2O3 -> 2Fe + 3H2O
Step 3: balance the H+ , on the  RHS, you see that there are 6H+, hence, to balance this, add 6H+ on the LHS
Fe2O3 + 6H+ -> 2Fe + 3H2O
Step 4: balancing all the charges by adding electrons. As you can see, on the LHS:
Fe2O3 + 6H+ -> 2Fe + 3H2O
(   0     +   6   ->  0    +  0     )      ( This step is actually finding the Oxidation number of each compound, add them together and find out where you can put the electrons to balance both side of the equation)
Hence, I see that there are 6+ on the LHS and 0 on the RHS, I would add 6e- to the LHS to balance the charges since 6 + (-6) = 0 ( = RHS)

Fe2O3 + 6H+ 6e- -> 2Fe + 3H2O
Step 5: include stages
Fe2O3(s) + 6H+(aq) + 6e- -> 2Fe(l) + 3H2O(l)
Hope this helps, let me know if there is any errors!

Firstly, where does the H+ come from? In the given question you just have aluminium and iron oxide. That's why my answer has an oxide in it.

I asked my teacher about this and he said that the 3O² isn't supposed to be there and now I'm pretty confused. I was wondering if someone could show me step by step to find this half equation.

Btw, if this is a reduction half equation, why aren't the electrons on the left side?

Thanks :-)

What does your teacher say? The products of the reaction should be aluminium oxide and molten iron metal.
And yeah it's a reduction except I'm stupid and I put the electrons on the wrong side of the equation. ALWAYS balance your charges. Not like what I'm doing here. I'm getting so careless now.

[nhmn0301]

Hmm I get it now, just find out the question is 2 pages away. Thanks anw!

[Blondie21]

What does your teacher say?

He said the answer should be what nhmn0301 stated 

[lzxnl]

He said the answer should be what nhmn0301 stated 

With waters? Ask him where the hydrogen atoms come from. The reaction involves aluminium metal with iron oxide, all in solid form.

[darklight]

Chloroform's molecules are made up of C, H and Cl. In an experiment, all of the chlorine present in a 3.59 gram sample of chloroform was converted into 12.92 gram of solid silver chloride, while the carbon content  of chloroform sample was converted into methanoic acid of mass 1.38g. 6.0 *10^23 chloroform molecules have the same mass as 6.0 *10^24 carbon atoms. Find the molecular formula of chloroform.

I know using the information that molar mass is 120 and the moles of C and Cl. But how do you find mole of H? Thanks!

[lzxnl]

You can work out the mass of chlorine and carbon in the sample of chloroform given. Subtract this from the mass of chloroform given, and you should have the mass of hydrogen in chloroform.

[nhmn0301]

He said the answer should be what nhmn0301 stated 

My teacher said what lzxnl stated. It actually makes sense to me now !

[Blondie21]

My teacher said what lzxnl stated. It actually makes sense to me now !

Do you mind explaining it to me? I still don't understand it lol

[jgoudie]

This is the main confusing part here:
m(C+H+Cl) = 3.59 (chloroform contains carbon hydrogen and chlorine)
m(H) = 3.95 - m(C + Cl)     (hydrogen is chloroform minus carbon and chlorine)

------ From the start:

The idea is to find the mole of chlorine from the silver chloride (AgCl) and carbon from methanol acid (CHOOH).

Convert this mole of chlorine and carbon to mass and work out the hydrogen via the following:

m(C+H+Cl) = 3.59 (chloroform contains carbon hydrogen and chlorine)
m(H) = 3.95 - m(C + Cl)     (hydrogen is chloroform minus carbon and chlorine)

Convert the m(H) into mole and then do you normal empirical formula of n(C:H:Cl)
Then find the molecular formula.

Hope this helps out.

Do you mind explaining it to me? I still don't understand it lol

[nhmn0301]

Do you mind explaining it to me? I still don't understand it lol

Sure, but since Im still trying to learn the concept, so please excuse me if I make any mistakes or get you confused:
I think the first thing that makes both of us make mistakes in this question is because we assume that, ALL redox reaction equations can be derived by adding water, H+ and stuff to both side without knowing when and where can we apply this concept.
Firstly, it is good to have a general idea of what is happening in a thermite process, the Fe2O3 and Al reacts together and generates extreme heat, this process is a redox reaction.
Before looking into the question, I guess you can write an overall equation for this:
Fe2O3 + Al -> Al203 + Fe  (sorry I'm really lazy to include states)
You can see that: Fe goes from +3 to 0 and Al goes from 0 to +3. Hence, Fe undergoes reduction and Al undergoes oxidation. Oxygen is actually just a spectator ions, do not even involve much in the redox process.
In Fe2O3, you know that you have 2Fe(3+) and 3O (-2), hence, the decomposition reaction comes as:
Fe2O3 -> 2Fe + 3O(-2)         If you question why at this stage, "2Fe" in the product has a charge of 0 but not +3, is because Fe has already gained 3 electrons from Aluminium, hence, it has successfully changed its charge to 0.
But we won't stop at that, since the equation is not balancing between the LHS and the RHS, since 3O (2-) carries a charge of -6, we need to add 6e- to the LHS as well. ( We are not adding H+ to balance because  in this reaction, there's no way water can present, we are talking about solid substances producing extreme heat). So our final answer should be:
Fe2O3 + 6e -> 2Fe + 3O(-2)
I know my explanation is not good enough and might still contain some errors, but it's my best, hope this helps!


 

[lzxnl]

Yeah, the chemistry is all there

[noah the lettuce]

I have this chemistry question that I'm not too sure about the answer, if anyone could help me out?
A certain mineral sample was found to contain some aluminium oxide, Al2O3. The Al3+ ions in this mineral can be selectively extracted by precipitating it as Al(C9H6NO)3. When a 2.600g sample of the mineral was treated this way, the precipitate weighed 0.1747g. What was the percentage by mass of the Al in the original ore?
Mineral=2.600g
m (Al(C9H6NO)3)= 0.1747g
M=459g/mol
n= 0.00038061mol=n (Al)

Al2O3--> 2Al
n (Al2)=1/2*n (Al)
n (Al2)= 0.000190305mol
M (Al2)= 54g/mol
m (Al2)=0.01027647g

% Al=(0.01027647/2.600)*100
        =0.3952%

[scribble]

what you've done is right, but the second part is a tad unecessry.
once you have n(Al)= 0.00038061mol
you can just go m(Al)=nM= 0.00038061*27.0 = 0.01027647g
and then workout the percentage instead of working out the mol of Al2 and then the molar mass of Al2

[noah the lettuce]

what you've done is right, but the second part is a tad unecessry.
once you have n(Al)= 0.00038061mol
you can just go m(Al)=nM= 0.00038061*27.0 = 0.01027647g
and then workout the percentage instead of working out the mol of Al2 and then the molar mass of Al2

Awesome, thank you  didn't realise I could skip that Al2 part, that cuts down the working out, but it's going to be a habit I'll have to cut out