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Suppose 8.00 g of CH4 is allowed to burn in the presence of 6.00 g of oxygen. How much (in grams) CH4, O2, CO2, and H2O remain after the reaction is complete?

All you chemistry freaks out there solve this one for me, and show all your working

If a 4x4 is travelling from Sharjah to Dubai on SZR at 8am on a Monday and a taxi is travelling from Dubai...

oh well.. let's see if I remember any of my A level Chemistry..

Hydrogen has weight 1, Carbon 4, Oxygen 2

Methane burns in Oxygen into Carbon Dioxide and water.

ch4 + 2 x(o2) = co2 + 2 x(h2o)

Therefore, if my logic and memory is correct 6g of Oxygen will react with 6g of methane to produce 6g of carbon dioxide and 6g of water, leaving 2g of Methane unreacted.

....

But then again, I could be totally wrong - I am an actuary by profession and don't deal with chemicals at all!!

1+1 = 2

But 2+2 = 4....

The world is round... But the universe is an ever expanding oval mass...

It is sunny today.

shafique wrote:If a 4x4 is travelling from Sharjah to Dubai on SZR at 8am on a Monday and a taxi is travelling from Dubai...

oh well.. let's see if I remember any of my A level Chemistry..

Hydrogen has weight 1, Carbon 4, Oxygen 2

Methane burns in Oxygen into Carbon Dioxide and water.

ch4 + 2 x(o2) = co2 + 2 x(h2o)

Therefore, if my logic and memory is correct 6g of Oxygen will react with 6g of methane to produce 6g of carbon dioxide and 6g of water, leaving 2g of Methane unreacted.

....

But then again, I could be totally wrong - I am an actuary by profession and don't deal with chemicals at all!!

hey, this sounds interesting.... chemistry lessons back in school..

shafique, you are partially incorrect. molecular weight of carbon is 12, hydrogen is 2 and oxygen 32.

if you balance the equation, you get one mole of methane reacting with two moles of oxygen to give one mole of carbondioxide and two moles of water.

if you put down the molecular weights in place, it means,

16 grams of methane reacts with 64 grams of oxygen to give 44 grams of carbondioxide and 36 grams of water.

find out the rest by maths, its simple...

it means that in the presence of 6 grams of oxygen, only 1.5 grams of methane will burn and the rest 6.5 grams will remain unreacted. this will produce 4.125 grams of carbondioxide and 3.375 grams of water.

the above looks to be simple, but actually there could be some other reactions taking place, a bit of carbon-monoxide may form because of lack of sufficient oxygen.. the calculation would become difficult then..

dont bother much on this....

Shafiq,

The values you mentioned were the number of free electrons in the last orbit, we use them for balancing the equation which ajoy did later.


Ajoy, you are using the right concept.

for the sake of the solution, just get the molaric weight and use the balanced equation down to do the calculation.

one mole of Methane, with 2 moles of oxygen get one mole of Co2 and 2 moles of water.

the difference in weight is lost as energy, if any

I last did any Chemistry 19 years ago!

You guys are right, I could not remember the molecular weights and got that confused with the no. of electrons... (valence?)....

I'm just glad I remembered as much as I did

Cheers,
Shafique