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Originally Posted by hasnainbk

if 16 grams of oxygen react with 28 gram of CO;
4.8g of O2 will combine with 8.4g of CO;
2CO(g) + O2(g) -> 2CO2(g)
Because 28 g of CO volumes 22.4 litres
8.4 gram will occupy 6.72
producing 6.48 litres of CO2.

==> thats the correct solution.

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This is incorrect, the ideal gas law (PV = nRT) says volume is proportional to n (number of molecules) not the mass.

Nothing Happened Here

thanks for the chance.

yep..it's n not m..but equation is right...as i demonstrated 1st..
ha

thats the way of converting grams into liters which didn't really mattered with the solution.you are right about this formula and it implements the same as I did.

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Originally Posted by zaa730

hmm....... O reely? can someone verify this?

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further more this:

Quote:

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2CO(g) + O2(g) -> 2CO2(g)
Because 28 g of CO volumes 22.4 litres

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is wrong because this

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2CO(g) + O2(g) -> 2CO2(g)

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equation references amount (ie moles) not mass therefore it takes 2 moles of CO for every 1 mol of O2. Also the question asked for the amount of gas in Liters and not just an equation

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How many liters of carbon monoxide (at STP) are needed to react with 4.8g of oxygen gas to produce carbon dioxide?

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Do you realized that you ask a bad/wrong question? Read it again.
Any positive number is a correct ans to that question.

Quote:

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Originally Posted by TP635

Quote:

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How many liters of carbon monoxide (at STP) are needed to react with 4.8g of oxygen gas to produce carbon dioxide?

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Do you realized that you ask a bad/wrong question? Read it again.
Any positive number is a correct ans to that question.

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The question is correct and has only one right answer if you interpret it as:
How many liters of carbon monoxide (at STP) are needed to fully react with 4.8g of oxygen gas such that there is none left over to produce carbon dioxide?

zaa730: I guess wrUmm568 deserves it more.He has really done some hard work.I am not sure if he is right or wrong but I appreciate him and would ask you to cancel the invite you sent to me and send it again to wrUmm568.

Quote:

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Originally Posted by wrUmm568

Quote:

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Originally Posted by TP635

Do you realized that you ask a bad/wrong question? Read it again.
Any positive number is a correct ans to that question.

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The question is correct and has only one right answer if you interpret it as:
How many liters of carbon monoxide (at STP) are needed to fully react with 4.8g of oxygen gas such that there is none left over to produce carbon dioxide?

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The original question is ambiguous. The CORRECT question is what you stated above. You should be given the invite for getting the right question and ans.

How many liters of carbon monoxide (at STP) are needed to react with 4.8g of oxygen gas to produce carbon dioxide?

2CO (g) + O2 (g) <=> 2CO2

Mole of O2 = 4.8 g/ (16 g/mol ) = # of Moles of O2

Moles of CO = 2 * (# of Moles of O2)

Then, you can go to liters simply by multiplying # of Moles of CO with 22.4 liters (coz the reaction is at STP)

or use classic PV = nTR

=> V = nTR / P

where,

P= pressure (1 atm at STP)

V = Volume

n = # of moles

T = Temperature in kelvin (0 degree celcius in STP)

R = constant (forgot the value lol)

P.S. me not want an invite...was barely showing http://www.pnwriders.com/forum/image...ies/epenis.gif