combustion

[logistic_guy]

here is the question

If $\displaystyle 1.5$ mol $\displaystyle C_2H_5OH$, $\displaystyle 1.5$ mol $\displaystyle C_3H_8$, and $\displaystyle 1.5$ mol $\displaystyle CH_3CH_2COCH_3$ are completely combusted in oxygen, which produces the largest number of moles of $\displaystyle H_2O$? Which produces the least? Explain.


my attemb
to solve this question, i need to do some calculations in the balanced chemical equation for combustion
if i do that the reaction only get one mole for each organic compound
how to do it for combustion of $\displaystyle 1.5$ mole?

 

[khansaheb]

here is the question

If $\displaystyle 1.5$ mol $\displaystyle C_2H_5OH$, $\displaystyle 1.5$ mol $\displaystyle C_3H_8$, and $\displaystyle 1.5$ mol $\displaystyle CH_3CH_2COCH_3$ are completely combusted in oxygen, which produces the largest number of moles of $\displaystyle H_2O$? Which produces the least? Explain.


my attemb
to solve this question, i need to do some calculations in the balanced chemical equation for combustion
if i do that the reaction only get one mole for each organic compound
how to do it for combustion of $\displaystyle 1.5$ mole?

Write three equations to show combustion reactions for the three organic compounds. Like:

2H₂+ O₂ = 2H₂O

 

[topsquark]

here is the question

If $\displaystyle 1.5$ mol $\displaystyle C_2H_5OH$, $\displaystyle 1.5$ mol $\displaystyle C_3H_8$, and $\displaystyle 1.5$ mol $\displaystyle CH_3CH_2COCH_3$ are completely combusted in oxygen, which produces the largest number of moles of $\displaystyle H_2O$? Which produces the least? Explain.


my attemb
to solve this question, i need to do some calculations in the balanced chemical equation for combustion
if i do that the reaction only get one mole for each organic compound
how to do it for combustion of $\displaystyle 1.5$ mole?

Stoichiometry. Basic Chemistry. Apparently another field you need to review.

Your equation for $CH_3CH_2COCH_3$ is wrong.

If I gave you 1 mol $C_2H_5OH$ how many mol $H_2O$ are produced?

Hint: Keep it in that ratio when you use 1.5 mol.

-Dan

 

[khansaheb]

Your equation for CH3CH2COCH3CH_3CH_2COCH_3CH3CH2COCH3

I do not see any equation in the OP!!

 

[topsquark]

I do not see any equation in the OP!!

He said the number of moles required is the same for each combustion. They are not. Ergo, he (at the least) screwed that one up.

-Dan

 

[logistic_guy]

Write three equations to show combustion reactions for the three organic compounds. Like:

2H₂+ O₂ = 2H₂O

khan you know chemistry. i'm surprise

it's better written like this
$\displaystyle 2H_2 + O_2 \to 2H_2O$

Stoichiometry. Basic Chemistry. Apparently another field you need to review.

don't worry it's in the top of my list

Your equation for $CH_3CH_2COCH_3$ is wrong.

then correct it

If I gave you 1 mol $C_2H_5OH$ how many mol $H_2O$ are produced?

reaction is my game, especially combustion
i'm chemical engineering and mechanical engineering, so this tell you i'm very good in reacions

let me play around in the balance eqaution
$\displaystyle C_2H_5OH + 3O_2 \to 2CO_2 + 3H_2O$

this is set for one mole, so for $\displaystyle 1.5$ mole i get

$\displaystyle 1.5(C_2H_5OH + 3O_2 \to 2CO_2 + 3H_2O) = 1.5C_2H_5OH + 4.5O_2 \to 3CO_2 + 4.5H_2O$

i think the answer to your question is $\displaystyle 4.5$ moles

is my analize correct?

 

[topsquark]

khan you know chemistry. i'm surprise

it's better written like this
$\displaystyle 2H_2 + O_2 \to 2H_2O$


don't worry it's in the top of my list


then correct it


reaction is my game, especially combustion
i'm chemical engineering and mechanical engineering, so this tell you i'm very good in reacions

let me play around in the balance eqaution
$\displaystyle C_2H_5OH + 3O_2 \to 2CO_2 + 3H_2O$

this is set for one mole, so for $\displaystyle 1.5$ mole i get

$\displaystyle 1.5(C_2H_5OH + 3O_2 \to 2CO_2 + 3H_2O) = 1.5C_2H_5OH + 4.5O_2 \to 3CO_2 + 4.5H_2O$

i think the answer to your question is $\displaystyle 4.5$ moles

is my analize correct?

Yes, but there is a much better way to approach this.
$\dfrac{1.5 \text{ mol } C_2H_5OH}{1} \times \dfrac{3 \text{ mol } H_2O}{1 \text{ mol } C_2H_5OH} = 4.5 \text{ mol } H_2O$

-Dan

 

[logistic_guy]

Yes, but there is a much better way to approach this.
$\dfrac{1.5 \text{ mol } C_2H_5OH}{1} \times \dfrac{3 \text{ mol } H_2O}{1 \text{ mol } C_2H_5OH} = 4.5 \text{ mol } H_2O$

-Dan

nice idea

Your equation for $CH_3CH_2COCH_3$ is wrong.

but you ignore to correct $\displaystyle CH_3CH_2COCH_3$

 

[topsquark]

nice idea


but you ignore to correct $\displaystyle CH_3CH_2COCH_3$

Nope. Balance the equation and apply the same rule as I did above. Surely this looks familiar...

-Dan

 

[logistic_guy]

Nope. Balance the equation and apply the same rule as I did above. Surely this looks familiar...

-Dan

Your equation for $CH_3CH_2COCH_3$ is wrong.

i'm sure you say my equation wrong
i bold it
read what you write before saying Nope

 

[topsquark]

i'm sure you say my equation wrong
i bold it
read what you write before saying Nope

Perhaps I read it wrong. So what's the combustion equation?

-Dan