Question

"Paraffin, a wax used to make candles, has a molecular formula of

C25H52 (s) + 38 O2 ---> 23 CO2 (g) + 26 H2O (l)

How many liters of Co2 are produced when a 23.4g paraffin candle burns at STP?"

Answer 1

Do you know what STP means? standard temp & pressure and at STP, one mole of gas occupies 22.4 L of volume (molar volume) = \(\sf 22~ \frac{mol}{L}\)

Answer 2

Take the mass given, conver to moles, and convert those moles of paraffin to CO\(\sf _2\)
keeping in mind that that it's a 1:23 ratio for paraffin: CO\(\sf _2\)
and finally, use STP to eliminate moles and get Liters.

Answer 3

I will start you off: 23.5 g paraffin \(\times\) \(\sf \frac{1~mol}{353.77~ g ~parrafin}\) = __________ mol of paraffin \(\sf \times \frac{23~mol~CO_2}{1~mol~parrafin}\) ...and you can finish it off from here.

Answer 4

Thanks for the help, abb0t! I'll try to figure it out. Don't have access to a calculator right now so idk if I can but thanks!

Answer 5

@abb0t Ok, I used my phone calculator & got 1.53, now what do I do?

Answer 6

You have moles of CO\(_2\) and you know at STP you have 22.7 mol/L....

Answer 7

What is the relationship here.

Answer 8

Your answer should be in the 10\(\sf ^{-2}\) values.

Answer 9

34.24?

Answer 10

Wrong.

Answer 11

I'm lost.

Answer 12

Do I divide 1.53/22.7? I'm sooo confused :(

Answer 13

Ok, so you have 1.53 moles of CO\(_2\), and at STP, you have 22.7 moles/L

it is the same as you did before, you cancel the units like in algebra: if you have: \(\sf \color{red}{\frac{a}{b} \times \frac{d}{a}}= \frac{d}{b}\) because the two \(\sf \color{red}{a}\)'s cancel out. Do you follow?
It is the same with units: \(\sf \color{red}{23.4~grams ~paraffin} \times \frac{1~mol}{\color{red}{353.77~g~paraffin}}\) = mol

Answer 14

\(\sf \color{red}{\cancel{23.4~grams ~paraffin}} \times \frac{1~mol}{\color{red}{{{\cancel{353.77~g~paraffin}}}}}\) = mol

Answer 15

Yeah... still isn't making sense to me. >.<

Answer 16

You're cancelling out the units. Because if you have: \(\sf \frac{grams~of~paraffin}{grams~of~paraffin} = 1 \) you're treating the units as constants or variables. If you have \(\sf \frac{4}{4} = 1\), same with \(\sf \color{red}{\frac{B}{B}} = 1\) AND SO FOrth.

Answer 17

That's not why I'm confused. It's just hard for me to understand when you don't use the actual values :x

Answer 18

I think I'm gonna take a break and come back to this problem or something..

Answer 19

\(\sf \color{red}{23.4~grams ~paraffin} \times \frac{1~mol~paraffin}{\color{red}{353.77~g~paraffin} }= \frac{\color{red}{23.4~g~paraffin}(1~mol~paraffin)}{\color{red}{353.77~g~paraffin}} = \frac{23.4~mol}{353.77 }\) = 0.0686 mol of paraffin

Answer 20

@abb0t so is .068 the answer? Cause that's what I got earlier but I thought I was wrong cause you said 10^-2 or something like that.

Answer 21

It is

Answer 22

No wait, it can't be cause I need Liters for the answer.

Answer 23

Then it should be 34.24

Answer 24

rite

Answer 25

He said no so idk I'm lost.

Answer 26

To get volume of one mole of any gas at STP , v got to multiply the moles by 22.4

Answer 27

ok I'll write that down then. thanks love :)