Confused with this part?

[Dencology]

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Hey guys:

so i still don't understand when to use O2 and when to use just O when i am calculating percent yeild and limiting reagent. i mean at some problem i see that 23gram is used fro calculating the moles of Oxygen and at other problems i see only 16grams is being used:

Ex:
water is formed by the addition of 4g of H2 to an excess of O2. if 27g of H2O is recovered, what is the percent yield for the reaction?

obviously, this is a simple problem so just writing the equation:

2H2 + O2 = 2H2O

4grams of H (1mole/2grams of H)(2moles of H2O/2 moles of H)(18g H2O/1mole of H2O)= 32grams of H20

Now: % yield= actual/ theoradical

(27/32) 100%= 75%

Now i thought that i would use just 1gram of Hydrgen instead of 2grams.

other problems we have lets say we have C:H:O. 45% of Carbon, 35% Hydrogen, 162gram sample. what is the molecular formula:

so we do 45 gC (1mole/12gram)= 3.7/1.25= approx= 3

35 gH (1mole/1gram)= 35/1.25=28

20 gO (1mole/16gram)= 1.25/1.25=1

then we divide everything by the smallest.

C3H28O1 then 3*12= 36
1*28=28
1*16= 16

total= 80

162/80=2
so the molecular formula is

C72H56O2.
Now why here we used 1 grams for hydrogen and not 2 grams of it for one mole of HYDROGEN. But in the first problem we used 1gram for one mole of Hydrogen and not 2grams?

thanks guys

 

[chessxwizard]

Try to do it step by step, maybe that will help you.

For the first one, it says we have 4g H2, right? Ok, so we find the moles of H2.

4/2=2 moles H2 to react with the O2 in excess (H2 is the limiting reagent, we are told).

2 H2 + O2 ==> 2 H20
This equation tells us that for every 2 moles of H2 and 1 mole of O2, we get 2 moles of H20. So... set up a fraction!

2 Moles H2 (how much is required) 2 Moles H20 (# moles theory prod.)
--------------------------------- = ----------------------------------
2 moles H2 (how much we have) X moles H20 theoretically produced

2x = 4; x = 2 moles of H20 theoretically produced.

Ok! so we have the # of moles of H20 produced. For every mole of H20, we know that the molecular weight is 18, so...

# of grams of H20 produced = 36g.

But from the actual reaction we know that only 27g are produced... so
27/36 x 100 = 75%

I think if you do it this way, you don't need to deal with questioning yourself if you need to use 2 or 1 mole of hydrogen in each step.

The second problem, if you use 2 grams for H2 you are already assuming that hydrogen is present within the compound in a multiple of 2... you can't do that. For those problems, just use the element's MW from the periodic table.