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Offline huskywolf

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stoichiometry volume produced
« on: December 06, 2009, 08:20:40 AM »
Hello
I have an important exam  coming up soon and I would like to make sure my calculations are correct

Q1.
Calculate volume of carbon dioxide produced at STP(22.4L) when 640g of Fe2O3 reduced to Fe
2Fe2O3 + 3C =4Fe + 3CO2

My answer = 134.4 mol L of CO2 produced
Could you tell me if this is the correct answer please

Offline cliverlong

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Re: stoichiometry volume produced
« Reply #1 on: December 06, 2009, 08:57:16 AM »
Can you show your method and calculations?

Often laying the detail out is a useful aid for one to check one's own work.

Clive

Offline huskywolf

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Re: stoichiometry volume produced
« Reply #2 on: December 06, 2009, 09:03:32 AM »
Calculate volume of carbon dioxide produced at STP(22.4L) when 640g of Fe2O3 reduced to Fe
2Fe2O3 + 3C =4Fe + 3CO2
n=m/M
so n(Fe2O3) = 640g/160gmol= 4mol Fe2O3
mole ratio= 4mol(fe2O30 x 3mol(CO2)/2mol(Fe2O3)=6 mol(CO2)
V=n.Vm
so
6mol(CO2) x 22.4L= 134 mol L


My answer = 134.4 mol L of CO2 produced

Thanks

Offline JEE10

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Re: stoichiometry volume produced
« Reply #3 on: December 06, 2009, 09:33:18 AM »
calculations are correct,

But you must remember to get your units correct.

volume of CO2 produced = n.Vm = 6 moles *22.4 (liter/mole) = 134.4 L and NOT 134.4 mol L

Offline huskywolf

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Re: stoichiometry volume produced
« Reply #4 on: December 06, 2009, 09:49:07 AM »
Hi yes sorry I forgot to cancel the L's


Offline huskywolf

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Re: stoichiometry volume produced
« Reply #5 on: December 06, 2009, 09:57:10 AM »
Q2.Find masses of H2O and CO2 produced when 32.2g  C12H22O11  is burned with excess oxygen

C12H22O11  + O2 = CO2 + H2O
C12H22O11  + 12O2=12CO2 + 11H2O
  n=m/M=> n(C12H22O11)=34.2g/342g mol-1= 0.1 mol  C12H22O11
 mole ratio
       0.1 mol(C12H22O11) x 12 mol(CO2)/1 mol(C12H22O11) = 1.2 moles CO2
       m=n.M
      m(CO2)=1.2mol x 12g mol-1= 14.4g CO2
                           (12O2 mol=12CO2mol)
         mole ratio
       0.1mol(C12H22O11) x 11mol (H2O)/1 mol(C12H22O11)= 1.1 moles H2O
       m=n.M
      m(H2O)=1.1 mol x 18g mol-1= 19.8g H2O

Offline huskywolf

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Re: stoichiometry volume produced
« Reply #6 on: December 06, 2009, 09:59:16 AM »
Could you please check my question 2
I am not sure if the amount of CO2 is correct
but both masses add up to original 32.2g

Offline sjb

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Re: stoichiometry volume produced
« Reply #7 on: December 06, 2009, 11:50:50 AM »
Could you please check my question 2
I am not sure if the amount of CO2 is correct
but both masses add up to original 32.2g

Not quite correct. If you have 32.2g C12H22O11 why did you calculate for 34.2g in
  n=m/M=> n(C12H22O11)=34.2g/342g mol-1= 0.1 mol C12H22O11

Offline huskywolf

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Re: stoichiometry volume produced
« Reply #8 on: December 06, 2009, 11:53:25 AM »
sorry its 34.2g (not 32.2g)

Now is it correct ? :D

Offline huskywolf

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Re: stoichiometry volume produced
« Reply #9 on: December 06, 2009, 11:54:35 AM »
original = 34.2g

Offline huskywolf

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Re: stoichiometry volume produced
« Reply #10 on: December 06, 2009, 01:35:24 PM »
Can anybody tell me if my Q2 is correct?????

Offline JEE10

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Re: stoichiometry volume produced
« Reply #11 on: December 07, 2009, 12:12:56 PM »
Q2.Find masses of H2O and CO2 produced when 32.2g  C12H22O11  is burned with excess oxygen

C12H22O11  + O2 = CO2 + H2O
C12H22O11  + 12O2=12CO2 + 11H2O
  n=m/M=> n(C12H22O11)=34.2g/342g mol-1= 0.1 mol  C12H22O11
 mole ratio
       0.1 mol(C12H22O11) x 12 mol(CO2)/1 mol(C12H22O11) = 1.2 moles CO2
       m=n.M
      m(CO2)=1.2mol x 12g mol-1= 14.4g CO2
                           (12O2 mol=12CO2mol)
         mole ratio
       0.1mol(C12H22O11) x 11mol (H2O)/1 mol(C12H22O11)= 1.1 moles H2O
       m=n.M
      m(H2O)=1.1 mol x 18g mol-1= 19.8g H2O

the approach and steps are correct.. just correct that 34.2 to 32.2 as has been mentioned and plug in the new values..

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