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Hudsonleaf · Nov 27, 2009, 12:03 PM

I am trying to Balance this Equation and having troubles.

I am supposed to put the co efficients in the spaces.

___MgF2 + Li2CO3=___MgCO3+___LiF

Here is also another one that I did but want to make sure I did it right?

___LiCl+___Br2=___LiBr+___Cl2

THANKS FOR YOUR HELP AHEAD OF TIME!! Very much appreciation~~~

InfoJunkie4Life · Nov 27, 2009, 07:17 PM

The object is to have the same amount of the same types of atoms on both side of the equations.

$MgF_2 + Li_2CO_3 = MgCO_3 + 2LiF$

InfoJunkie4Life · Nov 27, 2009, 07:22 PM

$2LiCl + Br_2 = 2LiBr + Cl_2$

mathwiz3502 · Nov 28, 2009, 04:29 PM

My teacher has us make a t chart with the number of atoms on each side, then we add coefficients and cross out and calculate the new number of atoms until they are equal.

InfoJunkie4Life · Nov 28, 2009, 10:43 PM

Excellent method, but I do most of it in my head. I've been working with these types of equations for years.

mathwiz3502 · Nov 29, 2009, 05:24 AM

So do I but some of the brightest students have to use this method.

InfoJunkie4Life · Nov 29, 2009, 09:28 AM

Oh, I understand.

Hudsonleaf · Nov 29, 2009, 12:37 PM

So what should go in the four spaces in order should be

2, 1, 2, 1 ?

Hudsonleaf · Nov 29, 2009, 12:44 PM

Here is one more, I would like to know if I did this one right also?

___PbBr2 +___ HCl=___HBr+___PbCl2

1 2 2 2??

Perito · Nov 29, 2009, 06:27 PM

$1 \, PbBr_2 + 2\, HCl \rightarrow 2\, HBr + 2\, PbCl_2\,\,\,???$

Almost, but it's not quite right. You have one lead atom on the left and two chlorine atoms on the left. On the right, you have two lead atoms and four chlorine atoms.

mathwiz3502 · Nov 30, 2009, 07:02 PM

[math]1PbBr_2+HCl/rightarrow2HBr+1PbCl_2

mathwiz3502 · Nov 30, 2009, 07:03 PM

Sorry,
$1PbBr_2+HCl/rightarrow2HBr+1PbCl_2$

mathwiz3502 · Nov 30, 2009, 07:06 PM

1221

Unknown008 · Dec 1, 2009, 07:03 AM

Originally Posted by Hudsonleaf

Here is one more, I would like to know if I did this one right also?

___PbBr2 +___ HCl=___HBr+___PbCl2

1 2 2 2????????????????

Ok, in your head, look at Pb first. There's one on both sides, so good.
Then, you see Br, there are two of them on the left and only one on the right. That means that you have to put '2' in from of HBr.

Doing so doubles the H. Since there's only one on the left hand side, double the HCl. That means you also have 2 Cl, but you already have 2 Cl on the left!

Ok, you saw everything. Time to check.
1 Pb on each side. Good
2 Br on each side. Good
2 H on each side. Good
2 Cl on each side. Good.

That's how I do, a sort of 'chaining' with the elements I'm dealing with, trying to fix one as soon as possible. That works for the easy equations. Other, more difficult require perseverance, and sometimes the use of half equations and fractions.