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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: huskywolf on December 06, 2009, 08:20:40 AM

Title: stoichiometry volume produced
Post by: huskywolf 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

Title: Re: stoichiometry volume produced
Post by: cliverlong 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

Title: Re: stoichiometry volume produced
Post by: huskywolf 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

Title: Re: stoichiometry volume produced
Post by: JEE10 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

Title: Re: stoichiometry volume produced
Post by: huskywolf on December 06, 2009, 09:49:07 AM

Hi yes sorry I forgot to cancel the L's

Title: Re: stoichiometry volume produced
Post by: huskywolf 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

Title: Re: stoichiometry volume produced
Post by: huskywolf 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

Title: Re: stoichiometry volume produced
Post by: sjb on December 06, 2009, 11:50:50 AM

Quote from: huskywolf 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

Not quite correct. If you have 32.2g C₁₂H₂₂O₁₁ why did you calculate for 34.2g in

Quote from: huskywolf on December 06, 2009, 09:57:10 AM

  n=m/M=> n(C₁₂H₂₂O₁₁)=34.2g/342g mol-1= 0.1 mol C₁₂H₂₂O₁₁

Title: Re: stoichiometry volume produced
Post by: huskywolf on December 06, 2009, 11:53:25 AM

sorry its 34.2g (not 32.2g)

Now is it correct ? :D

Title: Re: stoichiometry volume produced
Post by: huskywolf on December 06, 2009, 11:54:35 AM

original = 34.2g

Title: Re: stoichiometry volume produced
Post by: huskywolf on December 06, 2009, 01:35:24 PM

Can anybody tell me if my Q2 is correct?????

Title: Re: stoichiometry volume produced
Post by: JEE10 on December 07, 2009, 12:12:56 PM

Quote from: huskywolf 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

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