Chem Balancing and writing equaltions

[Burrin]

Arg Chem is blowing me out of the water. IF I ever want to be a doctor these grades are holding me back. Does any one have an hints for balancing chem equations...?

[jay7]

Give us an example, and I or someone else can step by step go through it. Then, after you practise about 300 of them, it will be natural. Much like solving for x in simple math equations..

I hated them at first as well.

[xXLoss_of_ControlXx]

can you post some problems?

its been a year since ive balanced equations but ill probably remember when i see it.

[Burrin]

Yeah I would love to but my teacher took all my work sheets to grade...eep... Well I will get one posted soon.

[Burrin]

Al+CuSO4---> Al2(SO4)3 + Cu

ANd


C3H8+ O2----> CO2 + H2O

[Iron Clad Ben]

Hey, I tutored chemistry (and some other stuff) in college.

Make sure you are treating compound anions like SO4 as one thing.

Here is my technique. Basically break the process up into chunks. Try to balance each element or compound ion one at a time. Look at the amount of each element on each side at each step. That will show you where you need to make adjustments.

Al+CuSO4---> Al2(SO4)3 + Cu

Taking inventory:
Left side: Al x 1, Cu x 1, SO4 x 1
Right side: Al x 2, Cu x 1, SO4 x 3

Which side has more of each thing? Both sides have the same amount of Cu. The right side of the equation/reaction has more SO4 and Al. So let's add some more Al and SO4 to the left side.

OK we need 2 Al's and 3 SO4s on the left side to even those out.

2Al+3CuSO4---> Al2(SO4)3 + Cu

Now we take inventory again:
Left side: Al x 2, Cu x 3, SO4 x 3
Right side: Al x 2, Cu x 1, SO4 x 3

Now the only difference is that the left side has more Cu. So we need to add more Cu to the right side.

2Al+3CuSO4---> Al2(SO4)3 + 3Cu

Final inventory / check:
Left side: Al x 2, Cu x 3, SO4 x 3
Right side: Al x 2, Cu x 3, SO4 x 3 it works!

------------------------------------

C3H8+ O2----> CO2 + H2O


Taking inventory:
Left side: C x 3, H x 8, O x 2
Right side: C x 1, H x 2, O x 3

This one looks tricky. Nothing initially matches. Let's balance C first. We need to add more to the right side (since the left side has 3 and the right side only has 1):

C3H8+ O2----> 3CO2 + H2O

Taking inventory again:
Left side: C x 3, H x 8, O x 2
Right side: C x 3, H x 2, O x 7

Now let's balance out the H's. There are more on the left, so let's add more to the right side

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

Taking inventory again:
Left side: C x 3, H x 8, O x 2
Right side: C x 3, H x 8, O x 10

Now finally we can balance out the O's to finish up. There are more on the right side, so let's add to the left side.

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

Final inventory / check:
Left side: C x 3, H x 8, O x 10
Right side: C x 3, H x 8, O x 10 it works!

[Burrin]

Wow thanks!!!!

[Iron Clad Ben]

Wow thanks!!!!

Did my post help you understand the process for solving them?

I tried to show the work step by step. Does it make sense? I didn't want to just post the answers and do your homework for you. That's no help to you in the long run.

Can you do these on your own now using the steps I showed? Chemistry problems, just like footbag, the way to get good is to PRACTICE. So keep doing lots of problems until you get super good at them and can do them in your sleep. Reading the book will only help you so much. PRACTICE PRACTICE PRACTICE.

If you run into any more that stump you post them up here, or PM me.

[Burrin]

Yeah it helps now we are on molar heat and stuff.

Here is one

Or two

When 2.8g of CaCls (s) dissolves in 20L of water, how much heat is released?
(DeltaHsoln= -82.8kj/mol)


ANd



Find the hea needed to melt 64.8g of copper at its melting point. ( The heat is Fussion of Copper is 13.38 kj/mol)

[Iron Clad Ben]

Hint: convert mass (grams) to moles. From there it's pretty simple.