[huskywolf]

Q2.Find masses of H2O and CO2 produced when 34.2g  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

Are my answers correct?(especially CO2)

[UG]

      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)

You've got the molar mass of CO₂ wrong. Apart from that, everything else looks fine.

[huskywolf]

I put in 12g on purpose ,I thought the two masses would need to add up to the original 34.2g??
I did not include the mass of 02 because the overall masses would exceed the 34.2g

[sjb]

I put in 12g on purpose ,I thought the two masses would need to add up to the original 34.2g??
I did not include the mass of 02 because the overall masses would exceed the 34.2g

As the oxygen is incorporated into your products (as per the balanced equation), where else is the source of the O₂ in CO₂? (or indeed H₂O) ?

[huskywolf]

Well if I put in the full mass for CO2 then the over mass of the two combined products would exceed 34.2g...
explain that to me!
I thought the mass of the products could not exceed the original!

[sjb]

The total mass of reactants, C₁₂H₂₂O₁₁ + O₂ is equal to the mass of products, CO₂ and H₂O

[Borek]

In other words: account for oxygen used for the reaction.