Calculate Mass of Butane & CO2 for 1.5x10^3 kJ Heat

[FLgirl]

1. What mass of butane in grams is necessary to produce 1.5 x 10^3 kJ of heat? What mass of CO2 is produced?

C4H10 + 13/2 O2 --> 4 CO2 + 5 H2O
Heat of reaction = -2658 kJ


2.



3. Mass of butane = 33g

Mass of CO2 = 176.04g <-- This is wrong and the right answer is 99g CO2





How is that so..

 

[halo31]

I think the reaction is exothermic. What it gives you is the overall delta H. In order to find what amount of butane is necessary to produce 1.5x10^3 kJ of energy(exothermic since its being released) is just a stochiometry problem. Here is a start.

-1.5x10^3 kJ(58.12g/-2658 kJ)=33 grams of Butane.

Now figure out how much CO2 is released.

 

[FLgirl]

I think the reaction is exothermic. What it gives you is the overall delta H. In order to find what amount of butane is necessary to produce 1.5x10^3 kJ of energy(exothermic since its being released) is just a stochiometry problem. Here is a start.

-1.5x10^3 kJ(58.12g/-2658 kJ)=33 grams of Butane.

Now figure out how much CO2 is released.

Thanks for answering... but why would you even need to know that to figure out how much CO2 was produced? Isn't it right there

 

[Borek]

176 g of CO₂ is produced from 1 mole of butane. 33 g of butane is not 1 mole.

 

[FLgirl]

176 g of CO₂ is produced from 1 mole of butane. 33 g of butane is not 1 mole.

Oh right! How would I solve it then?

 

[FLgirl]

33 / 58.12 = 0.567
176 x 0.567 = 99.7

 

[Borek]

Oh right! How would I solve it then?

Best approach is to try first, ask questions later.

33 / 58.12 = 0.567
176 x 0.567 = 99.7

Wasn't that hard.

 

[FLgirl]

Best approach is to try first, ask questions later.



Wasn't that hard.

Not really but if people were more inclined to solve the problem as an example, it would make learning go a lot quicker.

 

[Borek]

Not really but if people were more inclined to solve the problem as an example, it would make learning go a lot quicker.

Actually - not. The more problems you try to solve by yourself, the better your general solving skills will be. Otherwise you will learn how to deal with hundreds of known problems, but you will be still helplessly lost whenever facing a new one.

Unfortunately that's the way schools work these days - they teach how to reproduce the solution, not how to solve the problem.

If some of us know to solve the problem at first sight it is not because we did similar problem in the past, but because we know how to approach every problem. Believe it or not, but filling our template is a first step.

 

[FLgirl]

You are so knowledged. Tell me more tips to get good at chemistry.

 

[raychild]

How did you get 176g of CO?

176 g of CO₂ is produced from 1 mole of butane. 33 g of butane is not 1 mole.

I don't understand where the 176g of CO₂ came from...

 

[Borek]

I don't understand where the 176g of CO₂ came from...

Neither do I

176g of CO₂ is produced when you burn 1 mole of butane. Question doesn't ask about 1 mole of butane, so 176g is not a correct answer. It is a correct answer to some other question.

Actually for every number you can construct a problem that the number will be a correct answer.

 

[raychild]

I'm lost on how to solve this problem still...I got the 33g of butane but am not sure how to get to the 99g of CO2.

And I agree with you about working problems out on our own first in order to learn but I have been looking at this one for way too long!

 

[Borek]

Once you get 33 g of butane it is the most basic stoichiometry.

Convert 33 g to moles of butane, calculate how many moles of CO₂ are produced, convert back to mass.

Compare: http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations

 

[raychild]

Phew, thank you! I appreciate your help. It all makes sense now :)