When you have a reaction in which something was burned, it's a combustion reaction. A combustion reaction always produces (products) carbon dioxide and water. You have your reactants diethel ether and oxygen, thus your products will be carbon dioxide and water. Just write out the equation, and then balance. Once that is complete, you should be able to set-up the math and figure out how many grams of carbon dioxide were produced.

If you need additional help, let us know. Best of luck!

How to balance hydrocarbon combustion equations.
This was taught to me by a high school chem teacher.

The method is called CHO, because you balance the elements in that order,
Combustion products are CO_{2} and H_{2}O

so.

C_{4}H_{10}O + O_{2} --> CO_{2} + H_{2}O

1. First balance the carbons by the proper coefficient for CO_{2}, since I have 4 C in diethyl ether my coefficient is 4. CHO
C_{4}H_{10}O + O_{2} --> 4CO_{2} + H_{2}O

2. To balance the hydrogens on the left I need a coefficient of 5 on the right, since I have 2 H in each H_{2}O CHO
C_{4}H_{10}O + O_{2} --> 4CO_{2} + 5H_{2}O

3. This last step needs some concentration...
a) To balance the oxygen we add all the oxygen atoms on the RIGHT side of the equation
b) and subtract any oxygen in the compound burned on the left side ( the diethyl ether)
c) then divide by 2 because oxygen is diatomic, O_{2}.

a) 4(2) + 5(1) = 13
b) 13 - 1 = 12
c) 12 / 2 = 6 CHO
C_{4}H_{10}O + 6O_{2} --> 4CO_{2} + 5H_{2}

Note: sometimes you might get a fraction for the O_{2} coefficient, put the fraction in then multiply the entire equation (both sides) by 2. This works with even scary hydrocarbons like cinnamaldehyde, the essence necessary for good pumpkin pie - C_{6}H_{5}CH=CHCHO.

"Diethyl ether is burned in excess of oxygen to produce 2.40 moles of water. How many grams of carbon dioxide were produced?"

So now we have a basic stoichiometry problem with information of a substituent "given" and information of a substituent "wanted"

Basic steps
1. Balanced equation
2. Determine moles of "given"
3. Use molar ratio of (moles "wanted") / (moles "given"); from balanced equation
4. Convert moles "wanted" to information requested.

I noticed your post in another forum. If I'm not mistaken, this whole balancing equation thing is what has you troubled. This is part of a two posts I made to another member
...

Determine the number of atoms of each kind on each side. add coefficients one at a time to bring the number of each atom into balance. This is difficult to describe, so check you textbook and follow closely the examples.
...
You might start with this method. (I call it see-saw method)
Try this for

1. Check that all compounds are written correctly, and never change subscripts after that.

2. Starting with number of ions on each side
Reactant side Product side
1 Ca 3 Ca
2 Cl 1 Cl
3 Na 1 Na
1 PO4 2 PO4

to balance the calcium I need a coefficient of 3 for CaCl_{2}, I now have
Reactant side Product side
3 Ca 3 Ca
6 Cl 1 Cl
3 Na 1 Na
1 PO4 2 PO4

to balance the chlorine I need a coefficient of 6 for NaCl, I now have
Reactant side Product side
3 Ca 3 Ca
6 Cl 6 Cl
3 Na 6 Na
1 PO4 2 PO4

to balance the sodium I need a coefficient of 2 for Na_{3}PO_{4}, I now have
Reactant side Product side
3 Ca 3 Ca
6 Cl 6 Cl
6 Na 6 Na
2 PO4 2 PO4

I notice that the phosphate is now balanced. I'm done.
Take it slow and methodical. Good luck. As I said you will get better with practice.
Don't forget Step 1!

Hope this helps you Mary. If you have a particular equation you'd like us to help you with, let us know.

Chemistry Tutor

Sun, 2007-10-28 22:23

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## Re: Equations

ok first you need a balanced chemical equation

do you know what type of reaction this is ?

Andrea

Sun, 2007-10-28 22:42

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## Re: Equations

Okay, would that be C

_{4}H_{10}O + O_{2}-->.....?Andrea

Sun, 2007-10-28 23:09

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## Re: Equations

I'm not sure what the product would be...?

Anonymous (not verified)

Mon, 2007-10-29 05:24

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## Re: Equations

When you have a reaction in which something was burned, it's a combustion reaction. A combustion reaction always produces (products) carbon dioxide and water. You have your reactants diethel ether and oxygen, thus your products will be carbon dioxide and water. Just write out the equation, and then balance. Once that is complete, you should be able to set-up the math and figure out how many grams of carbon dioxide were produced.

If you need additional help, let us know. Best of luck!

Andrea

Mon, 2007-10-29 12:09

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## Re: Equations

Would the balanced equation be:

C

_{4}H_{10}+ O_{2}---> 5H_{2}O + 2CO_{2}?Andrea

Mon, 2007-10-29 12:11

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## Re: Equations

I mean:

C

_{4}H_{10}O + O_{2}---> 5H_{2}O + 2O_{2}?mr_cochran

Mon, 2007-10-29 13:10

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## Re: Equations

How to balance hydrocarbon combustion equations.

This was taught to me by a high school chem teacher.

The method is called CHO, because you balance the elements in that order,

Combustion products are CO

_{2}and H_{2}Oso.

C

_{4}H_{10}O + O_{2}--> CO_{2}+ H_{2}O1. First balance the carbons by the proper coefficient for CO

_{2}, since I have 4 C in diethyl ether my coefficient is 4.~~C~~HOC

_{4}H_{10}O + O_{2}--> 4CO_{2}+ H_{2}O2. To balance the hydrogens on the left I need a coefficient of 5 on the right, since I have 2 H in each H

_{2}O~~CH~~OC

_{4}H_{10}O + O_{2}--> 4CO_{2}+ 5H_{2}O3. This last step needs some concentration...

a) To balance the oxygen we add all the oxygen atoms on the RIGHT side of the equation

b) and subtract any oxygen in the compound burned on the left side ( the diethyl ether)

c) then divide by 2 because oxygen is diatomic, O

_{2}.a) 4(2) + 5(1) = 13

b) 13 - 1 = 12

c) 12 / 2 = 6

~~CHO~~C

_{4}H_{10}O + 6O_{2}--> 4CO_{2}+ 5H_{2}Note: sometimes you might get a fraction for the O

_{2}coefficient, put the fraction in then multiply the entire equation (both sides) by 2. This works with even scary hydrocarbons like cinnamaldehyde, the essence necessary for good pumpkin pie - C_{6}H_{5}CH=CHCHO.Andrea

Mon, 2007-10-29 13:52

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## Re: Equations

Oh....wow we were never taught this in class! Okay so now do I need to find the molecular weight? I'm not sure how to set up the math part...

mr_cochran

Mon, 2007-10-29 14:09

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## Re: Equations

"Diethyl ether is burned in excess of oxygen to produce 2.40 moles of water. How many grams of carbon dioxide were produced?"

So now we have a basic stoichiometry problem with information of a substituent "given" and information of a substituent "wanted"

Basic steps

1. Balanced equation

2. Determine moles of "given"

3. Use molar ratio of (moles "wanted") / (moles "given"); from balanced equation

4. Convert moles "wanted" to information requested.

Check your text for examples.

Hope this helps.

Andrea

Mon, 2007-10-29 23:20

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## Re: Equations

Got it, thanks so much!

Anonymous (not verified)

Fri, 2007-11-02 14:46

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## Re: Equations

i still dont get it >:( :(

mr_cochran

Fri, 2007-11-02 18:57

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## Re: Equations

Hello Mary,

What don't you get; which step.

1 Balancing an equation?

2 Converting a quantity to moles?

3 Using a molar ratio from the balanced equation?

4. Converting moles to another quantity?

Or does the whole thing seem too difficult?

We're here to help, but you need to let use know where the problems/confusions are.

mr_cochran

Fri, 2007-11-02 19:46

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## Re: Equations

Hello again Mary,

I noticed your post in another forum. If I'm not mistaken, this whole balancing equation thing is what has you troubled. This is part of a two posts I made to another member

...

Determine the number of atoms of each kind on each side. add coefficients one at a time to bring the number of each atom into balance. This is difficult to describe, so check you textbook and follow closely the examples.

...

You might start with this method. (I call it see-saw method)

Try this for

CaCl

_{2}+ Na_{3}PO_{4}--> Ca_{3}(PO_{4})_{2}+ NaCl1. Check that all compounds are written correctly, and never change subscripts after that.

2. Starting with number of ions on each side

Reactant side Product side

1 Ca 3 Ca

2 Cl 1 Cl

3 Na 1 Na

1 PO4 2 PO4

to balance the calcium I need a coefficient of 3 for CaCl

_{2}, I now haveReactant side Product side

3 Ca 3 Ca

6 Cl 1 Cl

3 Na 1 Na

1 PO4 2 PO4

to balance the chlorine I need a coefficient of 6 for NaCl, I now have

Reactant side Product side

3 Ca 3 Ca

6 Cl 6 Cl

3 Na 6 Na

1 PO4 2 PO4

to balance the sodium I need a coefficient of 2 for Na

_{3}PO_{4}, I now haveReactant side Product side

3 Ca 3 Ca

6 Cl 6 Cl

6 Na 6 Na

2 PO4 2 PO4

I notice that the phosphate is now balanced. I'm done.

Take it slow and methodical. Good luck. As I said you will get better with practice.

Don't forget Step 1!

Hope this helps you Mary. If you have a particular equation you'd like us to help you with, let us know.