chemistry hard question percentage yield

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usernamenew
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Magnesium reacts with oxygen as shown in the equation below: 2Mg + O2  2MgO Calculate the percentage yield of the reaction, given that burning 2.32 g of magnesium produced 2.39 g of magnesium oxide. (4 marks)
I thought %yield = theoretical yield/maximum yield so is that all you do
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Quady
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Yes
and is therefore not hard
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usernamenew
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(Original post by Quady)
Yes
and is therefore not hard
ok thanks was wondering though why it would be 4 marks if youre just dividing the 2 given numbers and then x100
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MexicanKeith
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(Original post by usernamenew)
ok thanks was wondering though why it would be 4 marks if youre just dividing the 2 given numbers and then x100
Most of the marks would be for working out theoretical yield.
you obviously have to convert from grams to moles and back again so I think 4 marks is fair enough
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Metanoia
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(Original post by usernamenew)
ok thanks was wondering though why it would be 4 marks if youre just dividing the 2 given numbers and then x100
In case you meant (2.32/2.39) x100%, that would be incorrect
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usernamenew
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(Original post by MexicanKeith)
Most of the marks would be for working out theoretical yield.
you obviously have to convert from grams to moles and back again so I think 4 marks is fair enough
oh i havent learnt that? i thought %yield was that simple equation thats why im confused because i dont know how you do the extra step, grams to moles?
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usernamenew
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(Original post by Metanoia)
In case you meant (2.32/2.39) x100%, that would be incorrect
yes that was what i meant.. lol , what am i supposed to do? ive only learnt that method
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MexicanKeith
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(Original post by usernamenew)
oh i havent learnt that? i thought %yield was that simple equation thats why im confused because i dont know how you do the extra step, grams to moles?
You should know an equation to convert between mass and moles

So, you can work out how many moles of magnesium you start with

you then know how many moles of MgO you can make

Then you can convert back again to get the theoretical yield of MgO in grams.

then you simply calculate [(actual yield)/ (theoretical yield)] * 100
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usernamenew
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(Original post by MexicanKeith)
You should know an equation to convert between mass and moles

So, you can work out how many moles of magnesium you start with

you then know how many moles of MgO you can make

Then you can convert back again to get the theoretical yield of MgO in grams.

then you simply calculate [(actual yield)/ (theoretical yield)] * 100
thanks but how do you know when you have to use moles in percentage yield because i havent come across that before?
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usernamenew
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(Original post by MexicanKeith)
You should know an equation to convert between mass and moles

So, you can work out how many moles of magnesium you start with

you then know how many moles of MgO you can make

Then you can convert back again to get the theoretical yield of MgO in grams.

then you simply calculate [(actual yield)/ (theoretical yield)] * 100
also is the answer 61.8%
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MexicanKeith
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(Original post by usernamenew)
thanks but how do you know when you have to use moles in percentage yield because i havent come across that before?
So you could use (actual mass/theoretical mass) *100

or (actual moles/theoretical moles)*100
both give you the same answer
Give me a second i'll work out the answer
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MexicanKeith
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(Original post by usernamenew)
also is the answer 61.8%
Yeah I make it around 62% (the exact value clearly depends on what values of molecular/ atomic masses you use)
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usernamenew
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(Original post by MexicanKeith)
Yeah I make it around 62% (the exact value clearly depends on what values of molecular/ atomic masses you use)
thank you so much for your help!
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