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i need one.
my reps are too low.
ahhh
well, if no one higher than 9 posts.... for a while, you'll be in luck
hold on....i'll give some stuff away real quick and get it up to 10.
:geptard:
i need TL invite.
Can you give it to me?
how can it be that +55 need a TL invite :) weird!
he is giving the inviteQuote:
Originally Posted by etrade
Today is opposite day. :P
good luck all i have acc + invite
good man and goodluck all
.
Nothing Happened Here
can I plz have it?
Answer:
2co+o2-> 2co2
2CO(g) O 2(g) y 2CO 2(g)
hey he copied my answer
nopes bosse de nage...look at your equation...you cant even write it correct.
Approximately 7.5708236 liters.
it's 3 liters for 6
3 meters?
oh well...i went to liberal arts college.
6.7 Liters:
2CO + O2 -> 2CO2
4.8g @ 32g/mol
0.30 0.15mol
PV = nRT -> V = nRT/P
STP: 1 atm, 273 K
or just use molar gas vol at 22.4 L /mol
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.
Nothing Happened Here
This is incorrect, the ideal gas law (PV = nRT) says volume is proportional to n (number of molecules) not the mass.Quote:
Originally Posted by hasnainbk
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.
further more this:Quote:
Originally Posted by zaa730
is wrong because thisQuote:
2CO(g) + O2(g) -> 2CO2(g)
Because 28 g of CO volumes 22.4 litres
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 equationQuote:
2CO(g) + O2(g) -> 2CO2(g)
Do you realized that you ask a bad/wrong question? Read it again.Quote:
How many liters of carbon monoxide (at STP) are needed to react with 4.8g of oxygen gas to produce carbon dioxide?
Any positive number is a correct ans to that question.
The question is correct and has only one right answer if you interpret it as:Quote:
Originally Posted by TP635
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.
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.Quote:
Originally Posted by wrUmm568The question is correct and has only one right answer if you interpret it as:Quote:
Originally Posted by TP635
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?
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
NOTHING HAPPENED HERE
kthxbye.
CLOSE TIME