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lgbasallote
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If 116.34 grams of butane reacts completely with O2 in STP conditions, what is the mass (grams) of CO2(g) formed?
do i use PV = nRT here?
 one year ago
 one year ago
lgbasallote Group Title
If 116.34 grams of butane reacts completely with O2 in STP conditions, what is the mass (grams) of CO2(g) formed? do i use PV = nRT here?
 one year ago
 one year ago

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GOODMAN Group TitleBest ResponseYou've already chosen the best response.0
Sorry, im taking chem next year.
 one year ago

GOODMAN Group TitleBest ResponseYou've already chosen the best response.0
@Mimi_x3 Help poor LG.
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
boohoo
 one year ago

GOODMAN Group TitleBest ResponseYou've already chosen the best response.0
I thought you were smart! You should know this! lol jk.
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
no im not. im dyslexic
 one year ago

Kryten Group TitleBest ResponseYou've already chosen the best response.2
i just woke up so lets start slowly.... :) first lets write down the equation of the reaction: 2 C4H10 + 13 O2 > 8 CO2 + 10 H2O now lets start: n=m/M n(C4H10)= 116,34 g/58,124 gmol1 = 2 mol n(CO2)/n(C4H10)= 8/2 = 4 n(CO2)= 4* n(C4H10) = 4* 2 mol = 8 mol m(CO2)= n(CO2)*M(CO2)= 8 mol * 44,009 gmol1 = 352,072 g
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
okay...how did you get that equation o.O
 one year ago

Kryten Group TitleBest ResponseYou've already chosen the best response.2
well it states that butane reacts with O2 so it is combustion of butane and it states that butane COMPLETELY reacts with O2 so it is combustion with infinite amount of O2 so reaction is as i wrote it...
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
so im guessing c4H10 is butane
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
so what happened after? in the "let's start"
 one year ago

Kryten Group TitleBest ResponseYou've already chosen the best response.2
you must learn if you already dont know homologous series: http://en.wikipedia.org/wiki/Homologous_series
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
well anyway the equation wasnt that important since it was a given :S
 one year ago

Kryten Group TitleBest ResponseYou've already chosen the best response.2
it is not what happened after lets start but what happened before it! in chemistry key to solving any task is to read few times what the task says and understand what is your goal... if you can see from the task what is your goal and your start line only thing you need to do is to find simpliest equation to come to your goal...
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
well okay that reaction was given...so now what?
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
we never studied homologous thingies so i doubt it'll appear in our quiz
 one year ago

Kryten Group TitleBest ResponseYou've already chosen the best response.2
ok so you have reaction, now you need to see what are you asked to do... (homologous series is general culture question more than chemistry question...) your task states that you have certain amount of reactant at certain conditions and that you need to calculate amount of product so you always start to calculate moles since moles are equivalent and masses are not! so you first calculate amount of that reactant in moles! get me so far?
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
hmm maybe a demonstration will
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
hmm sorry i just cant get this >.< maybe i'll come back to this once i have gained more experience...maybe then i can understand it...i'll try easier questions for now
 one year ago
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