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IGCSE CHEMISTRY DOUBTS HERE !!!!

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Vin:
Oh I'm really sorry. :-[ I'm afraid I don't know the 2nd one. :-[

1) a)

2C5H11OH + 15O2 ---> 10CO2 + 12H2O

Molecular mass of CO2 = 12 + 32
= 44 g

molar ratio of

    CO2 : O2  
    15 : 10
      1 : x

so, x = 10/15

Mass of CO2 = 10/15 * 44
= 29.3 g

b) Ratio of

  pentanol : CO2
          2 : 10
           1 : x
so for 1 mole, x = 5

Therefore 1 mole of pentanol = 5 moles of CO2 is formed
= 44 * 5
= 220 g

Now, you gotta find for 1 gram of pentanol

Mass of pentanol = 60 + 11 + 16 + 1
= 88g.

So,  
Pent.   CO2
88g      220g
1g         x

x = 220/88
= 2.5 g

$H00t!N& $t@r:
thanks alot for the detailed answer!  :) as for no. 2 i'll try to figure it out  ;)

Vin:

--- Quote from: Shooting Star on September 14, 2010, 07:51:28 pm ---thanks alot for the detailed answer!  :) as for no. 2 i'll try to figure it out  ;)

--- End quote ---

You're welcome. Hope you got it. It's a bit crude. I'll ask my teacher too tmw. ;)

Vin:

--- Quote from: Shooting Star on September 14, 2010, 07:10:32 pm ---yeah i do
2) 143g

--- End quote ---

Are you sure about the answer? Also is the question you know, complete?
It should be 80g according to me. :\

Deadly_king:

--- Quote ---2C5H11OH + 15O2 -----> 10CO2 + 12H2O

a) According to the equation :
15 moles of oxygen are required to produce 10 mole of carbon dioxide
15 moles of oxygen are required to produce (10 x44)g of carbon dioxide ( Mr of CO2 = 44)
Therefore 1 mole of oxygen will produce (10 x 44)/15 = 29.3g of CO2

b) According to the equation :
2 moles of pentanol burns to form 10 moles of carbon dioxide
(2 x 88)g of pentanol burns to form (10 x 44)g of carbon dioxide (Mr of pentanol = 83)
1g of pentanol will therefore burn to form (10 x 44) / (2 x 88) = 2.5g of CO2

2) 4Fe + 3O2 ----> 2Fe2O3

According to the equation :
For complete oxidation to take place, 4 moles of Fe reacts with 3 moles of O2 to form 2 moles of Fe2O3
1 mole of Fe reacts with 3/4 moles of O2 to form 1/2 moles of Fe2O3
1 mole of Fe reacts with 3/4 moles of O2 to form (1/2 x 159.6)g of Fe2O3
Mr of Fe2O3 = 159.6

Mass of iron (iii) oxide formed upon complete combustion of 1 mole of Fe = (159.6/2) = 79.8g
--- End quote ---

I agree with Percy.

Hey shooting star......u asked the same question here (https://studentforums.biz/sciences-149/chem-mole-calculation-plz-help/msg349416/#msg349416)

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