Hey guys, I'm stuck again with this question. I know how to work out the answer for the first question. But the second question seem complicated for me .. I got these answers from the mark scheme. Could any one explain for me why they multiplied 0.75 with 0.00893.
When heated, iron(III) nitrate (Mr = 241.8) is converted into iron(III) oxide,nitrogen dioxide and oxygen.4Fe(NO3)3(s) 2Fe2O3(s) + 12NO2(g) + 3O2(g) A 2.16 g sample of iron(III) nitrate was completely converted into the products shown.2 (a) (i) Calculate the amount, in moles, of iron(III) nitrate in the 2.16 g sample.Give your answer to 3 significant figures.[1 mark]
The answer for this one is 2.16 ÷ 241.8 = 0.00893
2 (a) (ii) Calculate the amount, in moles, of oxygen gas produced in this reaction.
The answer of this one is 0.00893 × 0.75 (= 0.00670 mol)
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- Thread Starter
- 23-02-2016 20:46
- 23-02-2016 22:01
The ratio of Iron(III) Nitrate to Oxygen is 4:3, which means one mole of nitrate will produce 3/4 moles of oxygen.
Therefore because you know the number of moles of Iron(III) Nitrate you can multiply this number by 3/4 (or 0.75) to get the number of moles of oxygen produced. (Because you'd divide both the reactant and the product by 4 to find one mole of it.)
- 23-02-2016 22:04
I'll try and keep my answer brief since well... there really isn't too much to say anyway!
Just to recap part a:
a) simply use the formula: number of moles = mass/Mr
put the numbers in and bang, there's the number of moles of iron(III) nitrate.
b) ok to understand this question you need to think about what the chemical equation shows you.
The numbers 4Fe(NO3)3(s) == 2Fe2O3(s) + 12NO2(g) + 3O2(g) shows the mole ratio (literally 4 moles of Iron(III) nitrate produces 2 moles of Fe2O3 plus 12 moles of 12NO2 plus 3 moles of O2).
Ok now lets tackle the question:
We know from part a that we've got 0.00893 moles of Iron.
What we do next to see how many moles of oxygen we have is to compare the ratio.
The ratio is 4:3 (from the numbers/coefficients)
This means that for every 4 moles of Fe(NO3)3, 3 moles of O2 is produced
But wait! We don't have 4 moles, we've got 0.00893.
So the question is really 'for every 0.00893 moles of Fe(NO3)3, how many moles of O2 is there?'
Simply divided 0.00893 by 4, then multiply it by 3 [which is the same thing as doing 3/4 of 0.00893].
Sorry I personally don't think this explanation is very good, but I hope i've helped a tiny bit!
Btw I'm an average A level student myself so approach my answer with scepticism!Last edited by Spectral; 23-02-2016 at 22:06.