Stoichiometry problem: Finding moles of product using mole-mole ratio of reactant & product

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

Thanks!

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ok first you need a balanced chemical equation

do you know what type of reaction this is ?

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Okay, would that be C4H10O + O2 -->.....?

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

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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!

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Would the balanced equation be:

C4H10 + O2 ---> 5H2O + 2CO2?

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I mean:

C4H10O + O2 ---> 5H2O + 2O2?

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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 CO2 and H2O

so.

C4H10O + O2 --> CO2 + H2O

1. First balance the carbons by the proper coefficient for CO2, since I have 4 C in diethyl ether my coefficient is 4.
CHO
C4H10O + O2 --> 4CO2 + H2O

2. To balance the hydrogens on the left I need a coefficient of 5 on the right, since I have 2 H in each H2O
CHO
C4H10O + O2 --> 4CO2 + 5H2O

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, O2.

a) 4(2) + 5(1) = 13
b) 13 - 1 = 12
c) 12 / 2 = 6
CHO
C4H10O + 6O2 --> 4CO2 + 5H2

Note: sometimes you might get a fraction for the O2 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 - C6H5CH=CHCHO.

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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...

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"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.

Hope this helps.

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Got it, thanks so much!

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i still dont get it  >:( :(

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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.

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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.
...
Try this for

CaCl2 + Na3PO4  -->  Ca3(PO4)2 + NaCl

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 CaCl2, 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 Na3PO4, 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.