Question

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?

Answer 1

Sorry, im taking chem next year.

Answer 2

@Mimi_x3 Help poor LG.

Answer 3

boohoo

Answer 4

I thought you were smart! You should know this! lol jk.

Answer 5

no im not. im dyslexic

Answer 6

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 gmol-1 = 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 gmol-1 = 352,072 g

Answer 7

okay...how did you get that equation o.O

Answer 8

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...

Answer 9

so im guessing c4H10 is butane

Answer 10

yep

Answer 11

so what happened after? in the "let's start"

Answer 12

you must learn if you already dont know homologous series:
http://en.wikipedia.org/wiki/Homologous_series

Answer 13

well anyway the equation wasnt that important since it was a given :S

Answer 14

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...

Answer 15

well okay that reaction was given...so now what?

Answer 16

we never studied homologous thingies so i doubt it'll appear in our quiz

Answer 17

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?

Answer 18

hmm maybe a demonstration will

Answer 19

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