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# mole question

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1. Calculate the mass of CO2 formed when 2.3 g of ethanol, C2H5OH reacts with 3.2 g oxygen O2 to form CO2 and H2O

C2H5OH + 3O2 → 2CO2 + 3H2O

I don't understand what should I do next cuz they gave us 2 masses
2. What do you mean, two masses? Of what?
3. (Original post by Hazel99)
C2H5OH + 3O2 → 2CO2 + 3H2OI am not sure why they gave u two masses maybe to combine masses 5.5/70= 0.08 (rounded) then times 2 =0.16 moles of CO2??Do u know the answer?
4.4g
4. C2H5OH mass no = 24 + 5 + 16 + 1 = 46

3O2 "mass" =3 X 2 x 16 = 96

46g ethanol reacts with 96g oxygen.

These masses are in an approx. ratio of ethanolxygen = 1:2, but the masses given (2.3g ethanol and 3.2g O2) are at a ratio of approx. 2:3. This tells you that there is an excess of ethanol. So the limiting factor is the oxygen amount. You use this to work out mass of CO2.

2CO2 "mass" = 2 X (12 + 32) = 88

96g of O2 would give 88g CO2

So 2.3g would give: 2.3/96 X 88
= 2.11g
5. 4.6x4--3.87= CO23HCL.9mg

Hope this helps!
6. (Original post by qatarownz)
Calculate the mass of CO2 formed when 2.3 g of ethanol, C2H5OH reacts with 3.2 g oxygen O2 to form CO2 and H2O

C2H5OH + 3O2 → 2CO2 + 3H2O

If we work out the number of mols of each of the reactants first we can see which is in excess:
Molar mass Ethanol = 46 g mol^-1
Molar mass 3O2 = 32 g mol^-1 * 3 = 96 g mol^-1 (notice the molar ratio)

Moles of Ethanol = 2.3g / 46 g mol^-1 = 0.05 mol
Mole of O2 = 3.2g / 96 g mol^-1 = 0.033(...) mol

Ethanol is in excess so we will use the Moles of O2 to calculate the final mass of CO2

Moles of CO2 = (0.033mol / 3) * 2 = 0.022 (notice the molar ratio)

Molar Mass of 2CO2 = 44 g mol^-1 * 2 = 88 g mol^-1 (notice the molar ratio)
Final Mass of CO2 = 0.022mol * 88g mol^-1 = 1.936g

(Original post by macpatelgh)
So 2.3g would give: 2.3/96 X 88
= 2.11g
The question says 3.2g O2, not 2.3g
7. Hi Thanks for pointing out my error - maybe I need to visit Specsavers (RE: ad ITV)

So 3.2g would give: 3.2/96 X 88= 2.93g
8. Unceleech's method (different approach from mine) is also correct and his maths seems ok, too; so why do I get a different answer? Would be grateful if you can work out my error, uncleleech, pls. (after I took into account the misreading of mass of oxygen!)
9. Hi again,

This was bugging me so I spent some time trying to spot the discrepancy between my logic and uncleleech's.

I notice uncleleech has taken into consideration the molar ratios of CO2 twice: firstly by a factor of 2/3 because of 2 moles of CO2 being yielded by 3 moles of O2, and secondly by a factor of 2 once again in the working out of the molecular mass of CO2. The definition of a mole is "the mass of a substance that is equivalent to the mass no expressed in grams".

Therefore, you cannot take effect of the molar ratios twice - I believe uncleleech has introduced an error of a factor of 2/3 by doing the first molar ratio manouvre. This introduced an error in his analysis of a factor of 2/3. His answer divided by my answer = 1.936/2.93 = 0.66 which is near enough to 2/3, considering we have both rounded off our answers.

It would be interesting to see the correct answer in any mark scheme.

Thanks
10. Okay I did it stupidly attached is my method:
11. It is not stupid at all! V easy error to make. Thanks again for you input and pointing out my my exotropia!
12. Just to anyone looking at this later:

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