Hard chemistry question GCSE! Help!

This discussion is closed.
#1
Iron is extracted by heating iron oxide with carbon monoxide.

Fe2 O3 + 3CO -----> 2Fe + 3CO2

a) Calculate the theoretical yield of iron that can be obtained when 320kg of iron oxide (Fe2 O3) reacts with excess carbon monoxide.

b) The actual yield of iron is 100kg. Calculate the percentage yield of iron.

This is really hard and not straight forward, I have got an answer that I have worked out, but not sure if it's right, could a really good chemist student work it out and see what you got? My answers are in the spoilers. Thanks!!

Spoiler:
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a) 146.9kg
b) 68.1%
0
4 years ago
#2
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0
4 years ago
#3
I got a different set of answers to you. This was my method:moles= mass / molecular mass so the number of moles of Fe2O3 is:320000g / 159.6 gmol-1 = 2005 molThen the ratio is 1:2 between the Fe2O3 and the Fe so you get 4010 mol of Fe. To get the mass you do 4010 x 55.8 = 223758g = 223.758kgWith this answer the percentage yield is 44.7%.
1
4 years ago
#4
(Original post by hannah-rosie)
I got a different set of answers to you. This was my method:moles= mass / molecular mass so the number of moles of Fe2O3 is:320000g / 159.6 gmol-1 = 2005 molThen the ratio is 1:2 between the Fe2O3 and the Fe so you get 4010 mol of Fe. To get the mass you do 4010 x 55.8 = 223758g = 223.758kgWith this answer the percentage yield is 44.7%.
i got the same answer. would it make a difference if u didn't convert from g to kg?
0
3 years ago
#5
I'm confused on how to get the answer.Can someone lay it out step by step?
0
3 years ago
#6
(Original post by Fergaljones)
I'm confused on how to get the answer.Can someone lay it out step by step?
1.First calculate the number of moles of Fe2O3:

The mass given is 320kg.
The formula mass is 2(56)+3(16) which is 160
The moles is mass/formula mass so 320/160 gives 2 moles.
2. From the reaction we see than 1 mole of Fe2O3 reacts with 2 moles of iron, so the moles of iron theoretically formed is 2*2 which is 4 moles. The mass is given by moles * formula mass which is 4*56 which is 224kg
3. The percentage yield is actual yield/ theoretical yield * 100 which is 100/224 * 100, giving 44.6 %
1
3 years ago
#7
Remember molar ratios and relevant formulae:

No. moles = Mass / Molar mass

The molar ratio of Fe2O3 to Fe is 1:2 - (1)Fe2O3, 2Fe.

There are 2000 moles of iron oxide (worked out with formula above), so we know by the ratio 1:2 that there muse be 4000 moles of iron in products, right?

Now work out the mass of 4000 moles of iron - that is the mass of iron you could theoretically expect to yield from 320kg of iron oxide.

Percentage yield = Actual yield / Theoretical yield.

You know your actual yield from the question, and you have worked out the theoretical yield.
Last edited by username4601382; 3 years ago
0
3 years ago
#8
yh thats def the right answer, easiest method as well.
(Original post by sangsan)
1.First calculate the number of moles of Fe2O3:

The mass given is 320kg.
The formula mass is 2(56)+3(16) which is 160
The moles is mass/formula mass so 320/160 gives 2 moles.
2. From the reaction we see than 1 mole of Fe2O3 reacts with 2 moles of iron, so the moles of iron theoretically formed is 2*2 which is 4 moles. The mass is given by moles * formula mass which is 4*56 which is 224kg
3. The percentage yield is actual yield/ theoretical yield * 100 which is 100/224 * 100, giving 44.6 %
0
3 years ago
#9
(Original post by sangsan)
1.First calculate the number of moles of Fe2O3:

The mass given is 320kg.
The formula mass is 2(56)+3(16) which is 160
The moles is mass/formula mass so 320/160 gives 2 moles.
2. From the reaction we see than 1 mole of Fe2O3 reacts with 2 moles of iron, so the moles of iron theoretically formed is 2*2 which is 4 moles. The mass is given by moles * formula mass which is 4*56 which is 224kg
3. The percentage yield is actual yield/ theoretical yield * 100 which is 100/224 * 100, giving 44.6 %
but 320 kg of Fe2O3 is 2000 mol of Fe2O3. You need to convert to g.
4000 mol of Fe is formed which is 224000 g or 224 kg
Last edited by username3249896; 3 years ago
0
3 years ago
#10
Okay thank you, I understand now.
(Original post by sangsan)
1.First calculate the number of moles of Fe2O3:

The mass given is 320kg.
The formula mass is 2(56)+3(16) which is 160
The moles is mass/formula mass so 320/160 gives 2 moles.
2. From the reaction we see than 1 mole of Fe2O3 reacts with 2 moles of iron, so the moles of iron theoretically formed is 2*2 which is 4 moles. The mass is given by moles * formula mass which is 4*56 which is 224kg
3. The percentage yield is actual yield/ theoretical yield * 100 which is 100/224 * 100, giving 44.6 %
0
2 years ago
#11
The molar mass of Fe2O3 = 2* 56 3*16 = 112 48 = 160 (g).=> Mole of Fe2O3 used = 320 000/ 160 = 2 000 (moles).=> Mole of Fe = 2 000 * 2 = 4 000 (moles).=> Theoretical yield = 4 000 * 56 = 224 000 (g) = 224 (kg).=> Percentage yield = Actual yield / Theoretical yield * 100 = 100/ 224 * 100 = 44.643 (%).
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