Who Can Balance Chemistry Equations?

[Appolo]

hey y'all! i was wondering if any of you can solve these equations. just fill in the blanks!



1.)_Fe2O3 + _CO = _Fe + CO2
2.)_Al2(SO4)3 + _Ca(OH)2 = _Al(OH)3 + _CaSO4
3.)_Na + _H2O = _NaOH + _H2
4.)_ Pb(CH3COO)2 + _H2S = _PbS + _CH3COOH
5.)_C5H12 + _O2 = _CO2 + _H2O

[Jack Squat]

Having trouble with homework?

EDIT: you forgot a space on that last one, there should be one right before the H2O

[A Rainy Knight]

I like the idea of this thread. I would normally do this kind of thing, but I don't have any paper with me right now. I'll edit in my answers in a little bit.

#3- 2, 2, 2, and 1.

[Solo]

1.) Fe2O3 + 3(CO) = 2Fe + 3(CO2)

2.)Al2(SO4)3 + 3(Ca(OH)2) = 2(Al(OH)3) + 3(CaSO4)

3.)2Na + 2(H2O) = 2(NaOH) + _H2

4.)_ Pb(CH3COO)2 + _H2S = _PbS + 2(CH3COOH)

5.)_C5H12 + 8(O2) = 5(CO2) + 6(H2O)

[Ryshan Ynrith]

Lessee here...
1. 1 + 3 = 2 + 3
2. 1 + 3= 2 + 3
3. 2 + 2 = 2 + 1
4. 1 + 1= 1 + 2
5. 1 + 8= 5 + 6

if I recall my chemistry properly.

An efficient algorithm is to make a matrix, with each equation representing the number of a particular element. For example, for number 5:

C: 5 n1=n3
H: 12 n1=2 n4
O: 2 n2=2 n3 +n4

Then solve for the variables, starting with n1, and use them as the coefficients of the compounds.

[Tam_OConnor]

Oh, balancing equations. This takes me way back...okay, so maybe it wasn't that long ago. Hope this doesn't violate the Paladin's code. Anyhow:

1) 2, 6 -> 4, 6
2) 1, 3 -> 2, 3
3) Concur with A Rainy Knight; 2, 2 -> 2, 1
4) 1, 1 -> 1, 2
5) 1, 8 -> 5, 6

May I suggest you get help from the teacher or classmates if you have trouble with these?

Edit: Curse those ninjas! Improved Uncanny Dodge! Improved Uncanny Dodge!

[Solo]

Your number one reduces to 1, 3, 2, 3.

[Appolo]

thank you guys a lot. now i have less homeworks and might be able to cath heroes tonight!

[Pyro]

Isn't balancing chemistry equations just making sure the number of lets say iron atoms and oxygen atoms are equal on both sides of the formula thingy?

[topher]

Isn't balancing chemistry equations just making sure the number of lets say iron atoms and oxygen atoms are equal on both sides of the formula thingy?

Exactly.

I was always required to show work though, hope you didn't forget that part Appolo.

[Appolo]

i dont have to show work. i just have to be sure my answers are right.

[Raiser Blade]

I need chemistry help please. I know there is a lot of problems but whatever you can help me with will be a lifesaver!


Balancing problems:

CO2(g) + H2(g) = H2O(g) + C(s)

HCI(aq) + AI(s) = AICI3(aq) +H2(g)

C10H20(1) + O2(g) = CO2(g) + H2O(g)

Word Problems:

Write a balanced equation for aqueous hydrogen bromide (HBr) reacting with solid calcium (Ca) to form aqueous calcium bromide (CaBr2) and gaseous hyrogen.

When C2H4 reacts with oxygen it forms carbon dioxide and water. Write out a balanced chemical equation.

When Lithium comes into contact with nitrogen and oxygen gas, it creates this compund LiNO2 Write out a balanced equation.


Calorimeter problems: (I hate these!)

If a 55.0 g glass marble lost 3251 Joules of heat from a starting temperature of 100 degrees celcius. What is it's new temperature?
(C of glass = .8372J/g *C

Complete the following calorimeter problem to find the specific heat of metal.

Data:
Calorimeter- 5.0 g, Ti =23.0 Celcius, c=2.34xJ/g C'
Water- 125.0 g, Ti = 23.0 Celcius, c=4.184xJ/g C
Metal- 36.0 g, Ti = 100.0 Celcius, c=?

T final for all substances =26.0 C


Any help would be greatly appreciated!

[Serpentine]

Balancing problems:

2CO2(g) + 2H2(g) = 4H2O(g) + 2C(s)

6HCI(aq) + 2AI(s) = 2AICI3(aq) +3H2(g)
(were those "I"s meant to be "l"?)

C10H20(1) + 15O2(g) = 10CO2(g) + 10H2O(g)

I think they're right, but I haven't done them for ages.

[The Extinguisher]

Can't do the Calorimeter, but I can do the rest.

CO2(g) + 2H2(g) -> 2H2O(g) + C(s)

6HCl(aq) + 3Al(s) -> 2AlCl3(aq) + 2H2(g)

C10H20](l) + 15O2(g) -> 10CO2(g) + 10H2O(g)

I won't do the word problems, cause you can't learn that way.

A quick and simple way I learned to balance. Keep a chart, or space for it. It makes life so much easier.
Other than that, you can't really show work for balancing.

Speaking of Chem Help, can someone help me with this? I can't for the life of me figure out what it makes.

Al(NO3)3 -> Al3+ + 3NO3-
3NO3- + 2H2O ->

It's driving me insane. I'm pretty sure it's an acid, but it might be a base.
Ugh.

[Raiser Blade]

(were those "I"s meant to be "l"?)


Yes and thanks for the help serp and FMA.


Anyone want to take a crack at those calorimeter problems? I'd be eternally grateful!

[Baboon Army]

Calorimeter problems: (I hate these!)

If a 55.0 g glass marble lost 3251 Joules of heat from a starting temperature of 100 degrees celcius. What is it's new temperature?
(C of glass = .8372J/g *C

Hmm lets see...I don't remember all the explanations, but there is a formula:
Q=m*c*ΔT.

Q is the energy, which is 3251j.
m is the mass, 55gr
and ΔT is the difference in temperature, and is equal to t-100˚C.

so what we get is:
Q=m*c*ΔT,
ΔT=Q/(m*c)
t-100=3251/(55*0.8372)
t=100+70.6 = 170.6˚C

Please correct me if I'm wrong.

EDIT:


I haven't done chem problems for eons, but if the marble lost heat, wouldn't its final temperature be lower than 100C, rather than higher?

Hmm...It could be -3251j, in this case we get 29.4˚C

seems right. I can't believe I did my chemistry exam only 6 months ago...I don't remember anything!

[Catch]

3NO3- + 2H2O ->

It's driving me insane. I'm pretty sure it's an acid, but it might be a base.
Ugh.

HNO3 + H2O → H3O+ + NO3-

"Nitric acid has an acid dissociation constant (pKa) of −1.4: in aqueous solution, it almost completely (93% at 0.1 M) ionizes into the nitrate ion NO3− and a hydrated proton, known as a hydronium ion, H3O+."

<3 Wikipedia

[EllysW]

I haven't done chem problems for eons, but if the marble lost heat, wouldn't its final temperature be lower than 100C, rather than higher?

[CrazedGoblin]

ive been doing chemistry for 3 years and i still cant balance equations weeeeeeeeeeeeeeeee:smallbiggrin:

[Thes Hunter]

:smallsigh: Balancing equations and the word problems and ect. all come from practice.


I hope Appolo and Raiser_Blade you guys are able to well on your up coming exams.



Though in the future it might be better to ask people how they go about solving the problems than asking just for the solutions. However, if that is all you want, you will reap what you sow.


Balancing equations is easy compared to other stuff that comes next. Standardized tests love to give the 'easy' questions on things like balancing equations. Just a warning.


However for the word problem... the HBr going to CaBR2 the first thing you should note is that there are two Br's in the final product. Start there and the rest of the balancing should become clear. And that is how I always balance my equations, I look for a 'keystone' element that has a certain proportion on the product of the equation, and then figure out how many it would take of the molecules on the reactants.

[Charity]

Isn't it basically all just algeba?

[Ryshan Ynrith]

If a 55.0 g glass marble lost 3251 Joules of heat from a starting temperature of 100 degrees celcius. What is it's new temperature?
(C of glass = .8372J/g *C




The equation to use is q=C(T2-T1). However, the C you are given has units of mass in it-it needs to be changed so it's merely J/C. Thus, divide the mass of the marble by your C value to obtain the C that enters into the equation. Note also that q is always assumed to be heat absorbed by the system; in this system it is negative because heat is lost. Thus, you have -3251=65.695(T2-100), which can be solved to find T2=50.5 C. If I remember correctly, that is.


Data:
Calorimeter- 5.0 g, Ti =23.0 Celcius, c=2.34xJ/g C'
Water- 125.0 g, Ti = 23.0 Celcius, c=4.184xJ/g C
Metal- 36.0 g, Ti = 100.0 Celcius, c=?
Tf=26

First off, we must find the energy in the system, which can be found by multiplying C by the mass and the initial temperature. This is problematic, as Ti does not have a C, but we can make two equations, one for the initial temperature and one for the final, and then solve them simultaneously.

Ei=12298.1+(36*100)Cti
Ef=13902.2+(36*26)Cti

Ef=Ei
13902.2+(36*26)Cti=12298.1+(36*100)Cti
1604.1=2664Cti
.60214 J/g*C=Cti

This is roughly analogous to listed values once unit conversions are made, so I am fairly confident in the answer.

[Thes Hunter]

Isn't it basically all just algeba?

Bah, algebra. We hates algebra!


I was balancing equations before I went back to re-learn algebra. But then again, for me personally the more intuitive approach with less math works better for me.


So maybe we are like a mirror image in some ways Charity me mate?

[LCR]

Isn't it basically all just algeba?

I refuse to believe that :smallwink:

[Raiser Blade]

Though in the future it might be better to ask people how they go about solving the problems than asking just for the solutions. However, if that is all you want, you will reap what you sow.


Hey i'll take what i can get! If you want to just explain how to solve the problems (especially the word ones) to me i'd be happy!

[Ryshan Ynrith]

A word of advice in any sort of physical science: watch your units. Always, always keep track of what units you are using, because you can be very sure that if you need a unit of length and end up with time, you've done something wrong.

My methodology is, hopefully, explained well enough above, but if you need a better explanation or more help, I'll be glad to give it. I'm a third-year Materials Science/Engineering major in college, so a lot of this stuff is still quite fresh. :)

[PlatinumJester]

I hate balancing those bastard equations :smallfurious:.


It NEVER makes sense :smallfurious:.

[Ryshan Ynrith]

The basic principal (of which I'm sure you're aware, I'm merely stating this for completeness) is that on each side of the reaction arrow, there must be the same amount of each element. Thus, in the combustion reaction C+O2->CO2, there is one carbon atom and two oxygen atoms on each side.

The mathematical way to accomplish this is to assign a variable to each term in the equation. For example, say this is the reaction:

C10H20(1) + O2(g) = CO2(g) + H2O(g)


Let n1 be the coefficient of the first term, n2 the coefficient of the second term, and so on.

Analyzing for carbon (C): 10n1=n3; that is, since there is only carbon in the first and third terms, and there is 10x as much carbon in the first as the third, that relationship must be true to balance the equation. Similarly, for hydrogen:

2n1=2n4

And finally, for oxygen:
n1+2n2=2n3+n4

To find each of them, solve all of them in terms of one. In this case, as n1 is involved in every equation, try to solve everything in terms of n1:

n3=10n1 from Carbon balance
n4=n1 from hydrogen balance

and plug them into the last equation:

n1+2n2=2(10n1)+n1

Solving this, we find that n2=10n1. n1 can then be declared as 1 (because there are no inconvenient fractions), and so the coefficients are such:

n1=1
n2=10
n3=10
n4=1

and the final, balanced equation is:

C10H20(1) + 10 O2(g) = 10 CO2(g) + H2O(g)

To check, merely sum the number of each element on each side:

Carbon on left: 1*10=10
Carbon on right: 10*1=10
Hydrogen on left: 1*2=2
Hydrogen on right: 1*2=2
Oxygen on left: 1*1+10*2=21
Oxygen on right: 10*20+1*1=21

Thus the equation is balanced. Does that make a little more sense? Any questions?

[sikyon]

Edit: Nvm, I found out

[Trog]

Wow. I still remember how to do these. *counts on fingers* 18 years later. :smalleek:

I was half my age when I last did these. :smallsigh:

[daggaz]

Im with everybody else who is warning you about getting quick answers off the net. You are gonna screw yourself, man.

If you arent too good at math and the algebra/matrix thing messes you up, there is another way to go about it.


Start with one kind of atom, balance it using the lowest possible numbers, ignore the other molecules. Then take another atom which is part of one of the molecules containing the first type of atom, and balance that, using lowest possible numbers. Continue. You might have to raise your the numbers of your original atom at some point.

If carbon is there, start with carbon, any molecule containing carbon on either side. Balance it. Then continue... Not as efficient as a pure math approach, but it will get you there.

For example, (sorry cant do subscripts)

_O2 + _C3H8 --> _CO2 + _H20

first carbon containing molecule has 3 carbons in it. So on the other side I will need 3 carbon dioxides to match one of those molecules (start with lowest possible numbers)

_O2 + 1C3H8 --> 3CO2 + _H2O

Now I could pick oxygen OR hydrogen. Oxygen is present in more than two molecules, so it is harder, I will save it for last. So i pick Hydrogen.

I have 8 Hydrogens in my first molecule (its a carbohydrate, propane to be exact), so I need 8 total in the H20. Each water molecule has two hydrogens each, so I need 4 of them to get a total of 8 hydrogen atoms. or..

_O2 + 1C3H8 --> 3CO2 + 4H2O

Now all that is left is oxygen, so I look at all the finished molecules on one side and count up the oxygen atoms there. I see six in the carbon-dioxide (three molecules with two oxygen each) and four from the water (four molecules with one oxygen each) making ten O atoms, so I need 10 oxygen atoms on the other side as well. O2, or atmospheric oxygen, has two oxygen atoms per molecule, so I need five molecules to hit ten O atoms.

5O2 + 1C3H8 --> 3CO2 + 4H2O

So five oxygen molecules interact with one propane to produce 3 carbon dioxides and 4 water molecules, plus a lot of heat. (heat not included).

That is a pretty basic situation, but you can get away with that approach for all but the most complicated problems, which I doubt you will have on your exam. Just remember, subscripts (the little hanging numbers after an atom) get multiplied by the superscript (the big number before the molecule) to give you the total number of atoms of one kind present in that kind of molecule. If you dont mess that up, it is hard to go wrong.

Good luck.

[Raiser Blade]

Im with everybody else who is warning you about getting quick answers off the net. You are gonna screw yourself, man.


But thanks to you and other helpful people not only am i getting answers i'm getting advice on how to solve similar problems.

Thanks to everyone who has helped so far! :smallsmile:

So could someone EXPLAIN how to go about writing out the word problems. I don't need the answer i'm just confused on what they want me to do exactly.

[Ryshan Ynrith]

These sorts of problems want you to set up an equation and balance it.

Take, for example, this one:

When C2H4 reacts with oxygen it forms carbon dioxide and water. Write out a balanced chemical equation.

You know that ethene is C2H4, and oxygen (being a gas) must be taken from the atmosphere as O2. Carbon dioxide is CO2, and of course water is H2O.

Thus, the unbalanced equation is:

C2H4 + O2 => CO2 + H2O

However, even on casual inspection this is unbalanced. Thus, using the algorithms and methods described above, you balance the equation. These questions are designed to test your knowledge of the formulas of common chemicals, recognizing reactants and products, and balancing equations. As a test, try balancing the equations here by yourself, without reference, then check your answers. It can seem very easy to do something when the way is detailed by someone else, but believe me, it gets a lot harder when there isn't any guidance at all.

Anyway, hope that helped.

[daggaz]

Word problems described.

Write a balanced equation for aqueous hydrogen bromide (HBr) reacting with solid calcium (Ca) to form aqueous calcium bromide (CaBr2) and gaseous hyrogen.


Ok, first in a word problem, underline the important objects, in this case the nouns and their adjectives (the chemicals and their states).

In problem one, we have HBr(aq), Ca(s), CaBr2(aq), H2(g). I know the hydrogen is in the H2 form because it said it was gaseous. You just have to know that. Now look at the verbs.

HBr(aq) reacts with Ca(s) to form CaBr2 and H2(g)

"Reacts with" is a + sign. "and" is a + sign. "to form" is an = sign, giving...

HBr(aq) + Ca(s) = CaBr2(aq) + H2(g)

Ok. Now we have an equation. The question asked us to balance it. So we pick an atom. Lets take Ca just because it sticks out, and it is only in two molecules, making it a good start.

_HBr(aq) + _Ca(s) = _CaBr2(aq) + _H2(g)

Using the start with the lowest number rule, we can see one calcium atom on each side, so they are already balanced! Yay!

_HBr(aq) + 1Ca(s) = 1CaBr2(aq) + _H2(g)

Now we move to Bromine, because it is the only atom in a molecule with Calcium. We have two bromines on the right side (2 atoms in one molecule), so we need a total of two on the left. Each HydrogenBromide molecule has only one Bromine atom, so we need two of those molecules, Or..

2HBr(aq) + 1Ca(s) = 1CaBr2(aq) + _H2(g)

Great, almost done. We need to balance hydrogen now. We have 2 on the left, we need two on the right. Each hydrogen gas molecule has two hydrogen atoms each, so we need only one gas molecule, or...

2HBr(aq) + 1Ca(s) = 1CaBr2(aq) + 1H2(g)

There, it is done. The (aq) and (s) and (g) only stand for which state the molecule is in, and turns out, are not at all important for this problem, tho later they will be, so get used to writing them down.

If you need help with the other two, PM me, Ill be up for a while.

[Raiser Blade]

Oh ok. Thanks a lot you guys! It's great that thee are nice people like you guys who will help a confused chemistry student such as myself.

I'm gonna try to do the other by myself but if i run into a snag i'll pm you daggaz. Thanks again for the help. :smallsmile:

[Raiser Blade]

When C2H4 reacts with oxygen it forms carbon dioxide and water. Write out a balanced chemical equation.


Is the answer:

C2H4 + 3O2 =2CO2 + 2H2O


@V Hows that.

[JT]

No, it isn't.

Remember that hydrogen, nitrogen, oxygen, and the "ine" sisters (flourine, chlorine, iodine, ...) all pair up when they aren't in a larger compound.

So it has to be __O2 on the left hand side.


(edit)
And... add up the number of oxygens you have on each side, and make sure they're equal.
(/edit)

[JT]

When C2H4 reacts with oxygen it forms carbon dioxide and water. Write out a balanced chemical equation.
Is the answer:
C2H4 + 3O2 =2CO2 + 2H2O

@V Hows that.

Much better. :)


But you should be able to verify it for yourself.
Add up the number of carbons atoms on each side. They should be equal.
Add up the number of hydrogen atoms. They should be equal.
Add up the number of oxygen atoms. They should be equal.
Check that you didn't miss any "gotchas" with atoms pairing up.
Check that you have an integer number of each molecule in the equation.

If those are all correct, then you almost have to be right!

[Raiser Blade]

When Lithium comes into contact with nitrogen and oxygen gas, it creates this compund LiNO2 Write out a balanced equation.

2Li + N2 + 2O2 = 2LiNO2

[JT]

"Remember that hydrogen, nitrogen, oxygen, and the "ine" sisters (flourine, chlorine, iodine, ...) all pair up when they aren't in a larger compound."

[Gaelbert]

When Lithium comes into contact with nitrogen and oxygen gas, it creates this compund LiNO2 Write out a balanced equation.

2Li + N2 + 2O2 = 2LiNO2

That one looks good to me. We just had to do these about a month ago. Next you get to look forward to moles and gram conversions if you're following the same path we are.

[Stormbright]

Tsk Tsk...
As a Teacher in the Playground, I must express a little displeasure with the wholesale providing of answers.... or seeking of them. While any teacher knows that homework may not always reflect authentic work, the PURPOSE is to prepare students for future concepts, and expected skills. What will you do on the test?

Just my 2 cents

[Raiser Blade]

What will you do on the test?

Just my 2 cents

What if that was my test. :smalltongue:

Just kidding actually though i got some good help from these guys. It wasn't just free answers.

[Ryshan Ynrith]

I have tried rather strenuously to teach the methods in lieu of merely listing answers-this is little different from going to a classmate or TA to get help with homework, a practice that is condoned and often encouraged, as it is not possible to learn with no basis for understanding.

I will say that if the people helped by this thread don't try to learn the methods we have described, they are doing themselves a grave disservice-as you say, what will you do on the test?-but I am hardly providing 'wholesale answers'.

[Stormbright]

actually, scanning back, Ryshan and Raiser do seem to have been somewhat responsible. Kudos. Admittedly, too tired to go back any further :-P